Chapter # 28
Heat Transfer
SOLVED EXAMPLES
EXAMPLE 28.1 One face of a copper cube of edge 10 cm is maintained at 1000 C and the opposite face is maintained at 00 C. All other surfaces are covered with an insulating material.Find the amount of heat flowing per second through the cube. Thermal conductivity of copper is 385 W/m– 0 C. Sol. The heat flows from the hotter face towards the colder face. The area of cross–section perpendicular to the heat flow is A = ( 10 cm ) 2 The amount of heat flowing per second is Q T1 T2 = KA t x
= ( 385 W/m–0C ) × ( 0.1 m ) 2 ×
(100 0 C 0 0 C ) 0.1 m
= 3850 W . EXAMPLE 28.2 Find the thermal resistance of an aluminimum rod of length 20 cm and area of cross–section 1 cm2 . The heat current is along the length of the rod . Thermal conductivity of aluminium = 200 W/m–K. Sol. The thermal resistance is R=
20 10 2 m x = = 10 K/W.. (200 W/m - K)(1 10 - 4 m 2 ) KA
EXAMPLE 28.3 The light from the sun is found to have a maximum intensity near the wavelength of 470 nm.Assuming that the surface of the sun emits as a blackbody, calculate the temperature of the surface of the sun. Sol.
For a blackbody, m T = 0.288 cm – K. Thus,
T=
0.288 cm - K = 6130 K. 470 nm
EXAMPLE 28.4 A blackbody of surface area 10 cm 2 is heated to 1270 C and is suspended in a room at temperature 27 C. Calculate the initial rate of loss of heat from the body to the room. 0
Sol.
For a blackbody at temperature T, the rate of emission is u = A T4 . When it is kept in a room at temperature T0 , the rate of absorption is u0 = A T04 . The net rate of loss of heat is u – u0 = A ( T4 – T40 ) . Here A = 10 × 10– 4 m 2 , T = 400 K and T0 = 300 K. Thus, u – u0 = ( 5.67 × 10– 8 W/m 2 – K 4 ) ( 10 × 10 – 4 m 2 ) ( 400 4 – 300 4 ) K 4 = 0.99 W.
EXAMPLE 28.5 A liquid cools from 700 C to 600 C in 5 minutes. Calculate the time taken by the liquid to cool from 600 C to 50 C , if the temperature of the surrdounding is constant at 300 C. 0
Sol. The average temperature of the liquid in the first case is
70 0 C 60 0 C = 65 0 C 2 The average temperature difference from the surrounding is 1 – 2 = 650 C – 300 C = 350 C. The rate of fall of temperature is 1=
–
d 70 0 C 60 0 C = = 2 0 C/min. dt 5 min
manishkumarphysics.in
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Chapter # 28 From Newton’s law of cooling,
Heat Transfer 20 C/min = bA ( 350 C )
2 35 min In the second case, the average temperature of the liquid is
or,
bA =
........(1)
60 0 C 50 0 C = 550 C 2 so that, 2 – 0 = 550 C – 300 C = 250 C. If it takes a time t to cool down from 600 C to 500 C, the rate of fall of temperature is 2 =
d 2 60 0 C 50 0 C 10 0 C = = dt t t From Newton’s law of cooling and (i),
–
2 10 0 C = × 250 C 35 min t t = 7 min.
or,
QUESTIONS 1. 2. 3. 4. 5. 6.
7. 8. 9. 10. 11.
FOR
SHORT
ANSWER
Q dQ . Why don’t we write ? t dt Does a body at 20ºC radiate in a room, where the room temperature is 30ºC ? If yes, why does its temperature not fall further ? Why does blowing over a spondful of hot tea cool it ? Does evaporation play a role ? Does radiation play a role ? On a hot summer day we want to cool our room by opening the refrigerator door and closing all the windows and doors. Will the process work ? On a cold winter night you are asked to sit on a chair. Would you like to choose a metal chair or a wooden chair ? Both are kept in the same lawn and are at the same temperature. Two identical metal balls one at T1 = 300 K and the other at T2 = 600 K are kept at a distance of 1m in vaccum. Will the temperature equalise by radiation ? Will the rate of heat gained by the colder sphere be proportional to T24 – T14 as may be expected from the Stefan’s law ? An ordinary electric fan does not cool the air, still it gives comfort in summer. Explain. The temperature of the atmosphere at a high altitude is around 500ºC. Yet an animal there would freeze to death and not boil. Explain. Starting in the sun is more pleasant on a cold winter day than standing in shade. Is the temperature of air in the sun considerably higher than that of the air in shade ? Cloudy nights are warmer than the nights with clean sky. Explain. Why is a white dress more comfortable than a dark dress in summer ?
The heat current is written as
Objective - I 1.
The thermal conductivity of a rod depends on (A) length (B) mass (C) area of cross-section (D*) material of the rod fdlh NM+ dh Å"ek pkydrk fuHkZj djrh gS (A) yEckbZ ij (B) nzO;eku ij (C) vuqiLz Fk dkV {ks=kQy ij (D*) NM+ ds inkFkZ ij
2.
In a room contaning air, heat can go from one place to another (A) by conduction only (B) by convection only (C) by radiation only
(D*) by all the three modes
ok;q ls Hkjs gq, dejs esa Å"ek ,d LFkku ls nwljs LFkku rd xeu dj ldrh gS (A) dsoy pkyu }kjk (B) dsoy laogu }kjk (C) dsoy fofdj.k }kjk (D*) rhuksa fof/k;ksa }kjk 3.
A sloid at temperature T1 is kept in an evacuated chamber at temperature T2 > T1. The rate of increase of temperature of the body is proportional to T1 rki okyh ,d Bksl oLrq] ,d fuokZfrr izdks"V esa j[kh gqbZ gS] ftldk rki T2 > T1 gSA oLrq ds rki esa o`f) dh nj fuEu ds lekuqikrh gS manishkumarphysics.in
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Chapter # 28 (A) T2 - T1
(B) T22 - T12
Heat Transfer (C) T23 - T13
(D*) T24 - T14
4.
The thermal radiation emitted by a body is proportional to Tn where T is its absolute temperature. The value of n is exactly 4 for (A) a blackbody (B*) all bodies (C) bodies painted balck only (D) polished bodies only fdlh oLrq ls mRlftZr gksus okys Å"eh; fofdj.k Tn ds lekuqikrh gS] tgk¡ T bldk ijerki gSA n dk eku Bhd 4 gksrk gS] fuEu ds fy;s (A) Ñf".kdk oLrq (B*) leLr oLrq,¡ (C) dsoy og oLrq,¡ ftu ij dkyk jax fd;k x;k gSA (D) dsoy ikWfy'k dh gqbZ oLrq,¡
5.
Two bodies A and B having equal surface area are maintained at temperatures 10oC. The thermal radiation emitted in a given time by A and B are in the ratio leku i`"Bh; {ks=kQy okyh nks oLrq,¡ A rFkk B 10oC o 20°C rki ij j[kh x;h Å"eh; fofdj.k dk vuqikr gksxk(A*) 1 : 1.15 (B) 1 : 2 (C) 1 : 4 (D) 1 : 16
6.
One end of a metal rod is kept in a furnace. In steady state, the temperature of the rod (A) increases (B) decreases (C) remains constant (D*) is nonuniform /kkrq dh ,d NM+ dk ,d fljk HkV~Vh esa j[kk gqvk gSA LFkk;h voLFkk esa NM+ dk rki (A) c<+sxk (B) de gksxk (C) fu;r jgsxk (D*) vle:i gksxk
7.
Newton’s law of cooling is a special case of (A) Wien’s displacement law (C*) Stefan’s law
(B) Kirchoff’s law (D) Planck’s law
U;wVu dk 'khryu dk fu;e] fuEu dh fo'ks"k fLFkfr gS (A) ohu dk foLFkkiu fu;e (B) fdpZgkWQ dk fu;e (C*) LVhQu dk fu;e (D) Iysd a dk fu;e 8.
A hot liquid is kept in a big room. Its temperature is plotted as a funcation of time. Which of the following curves may represent the plot ?
,d xeZ nzo fdlh cM+s dejs esa j[kk gqvk gSA bldk rki le; ds Qyu ds :i esa vkjsf[kr fd;k x;k gSA fuEu esa ls dkSulk xzkQ bldks O;Dr djrk gS -
(A*) a 9.
(B) c
(C) d
(D) b
A hot liquid is kept in a big room. The logarithm of the numerical value of the temperature difference between the liquid and the room is plotted against time. The plot will be very nearly (A*) a straight line (B) a circulat arc (C) a parbola (D) an ellipse
,d xeZ nzo fdlh cM+s dejs esa j[kk gqvk gSA nzo o dejs ds rkiksa ds varj ds ykWxjsf;e dk le; ds lkis{k xzkQ vkjsf[kr fd;k tkrk gSA xzkQ gksxk] yxHkx (A*) ljy js[kk (B) o`Ùkkdkj pki (C) ijoy; (D) nh?kZoÙ` k 10.
A body cools down from 65°C to 60°C in 5 minutes. It will cool down from 60°C to 55°C in (A) 5 minutes (B) less than 5 minutes (C*) more than 5 minutes (D) less than or more than 5 minutes depending on whether its mass is more than or less than 1 kg. ,d oLrq 5 fefuV esa 65°C ls 60°C rd B.Mh gksrh gSA ;g 60°C ls 55°C rd B.Mh gksxh (A) 5 fefuV (B) 5 fefuV ls de le; esa (C*) 5 fefuV ls vf/kd le; esa (D) 5 fefuV ls de ;k vf/kd le; es]a ;g bl ij fuHkZj djsxk fd oLrq dk nzO;eku 1 xzke ls vf/kd gS vFkok de
Objective - II 1.
One end of a metal rod is dipped in boiling water and the other is dipped in melting ice. (A) All parts of the rod in thermal equilibrium with each other (B) We can assign a temperature to the rod. (C) We can assign a temperature to the rod after steady state is reached manishkumarphysics.in
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gSA
Chapter # 28
Heat Transfer
(D*) The state of the rod does not change after steady state is reached.
/kkrq dh NM+ dk ,d fljk mcyrs gq, ikuh esa rFkk nwljk fljk fi?kyrs gq, cQZ esa Mqck;k x;k gS (A) NM+ ds leLr Hkkx ijLij Å"eh; lkE;koLFkk esa gksx a As (B) ge NM+ dk rki fu/kkZfjr dj ldrs gSAa (C) ge LFkk;h voLFkk izkIr gksus ds i'pkr~ NM+ dk rki fu/kkZfjr dj ldrs gSAa (D*) LFkk;h lkE;koLFkk izkIr gksus ds i'pkr~ NM+ dh voLFkk ifjofrZr ugha gksxhA 2.
A blackbody does not : (A) emit radiation (B) absorb radiation (C*) reflect radiation dksbZ Ñf".kdk oLrq ughs djrh gS (A) fofdj.k dk mRltZu (B) fofdj.k dk vo'kks"k.k (C*) fofdj.k dk ijkorZu (D*) fofdj.k dk ikjxeu
(D*) refract radiation
3.
In summer, a mild wind is often found on the shore of a calm river. This is caused due to (A) difference in thermal conductivity of water and soil (B*) convection currents (C) conduction between air and the soil (D) radiation from the soil xfeZ;ksa es]a 'kkar unh ds fdukjs ij 'khry ok;q izokfgr gksrh gSA bldk dkj.k gS (A) ikuh rFkk feV~Vh dh Å"ek pkydrk dk varj (B*) laogu /kkjk,¡ (C) gok rFkk feV~Vh ds chp pkyu (D) feV~Vh ls fofdj.k
4.
A piece of charcoal and a piece of shining steel of the same area are kept for long time in an open lawn in bright sun. (A) The steel will absorb more heat than the charcoal. (B) The temperatuer of the steel will be higher than that of the charcoal. (C*) If the both are picked up by bare hands, the steel will be felt hotter than the charcoal (D*) If the two are picked up from the lawn and kept in a cold chamber, the charcoal will lose heat at a faster rate than the steel
leku {ks=kQy okys ydM+h ds dks;ys ds VqdM+s rFkk pednkj LVhy ds VqdM+s dks pkqys ykWu esa pedrs lw;Z ds uhps yEcs le; rd j[kk tkrk gS (A) LVhy] dks;ys dh rqyuk esa vf/kd Å"ek vo'kksf"kr djsxkA (B) LVhy dk rki] dks;ys ds rki ls vf/kd gksxkA (C*) ;fn nksuksa dks uaxs gkFkksa ls mBk;k tk;s rks dks;ys dh rqyuk esa LVhy vf/kd xeZ yxsxkA (D*) ;fn nksuksa dks ykWu ls mBkdj fdlh B.Ms izdks"B esa j[k fn;k tk;s rks LVhy dh rqyuk esa dks;ys ls Å"ek gkfu vf/ kd rsth ls gksxhA 5.
A heated body emits radiation which has maximum intensity near the frequency v0. The emissivity of the material is 0.5. If the absolute temperature of the body is doubled (A*) the maximum intensity of radition will be near the frequency 2v0 (B) the maximum intensity of radiation will be near the frequency v0/2 (C*) the total energy emitted will increase by a factor of 16 (D) the total energy emitted will increase by a factor of 8 ,d xeZ dh x;h fofdj.k mRlftZr dj jgh gS] ftldh vf/kdre rhozrk v0 vko`fÙk ds lehi gSA inkFkZ dh mRltZdrk 0.5 gSA ;fn oLrq dk ijeki nqxuk dj fn;k tk;s (A*) fofdj.k dh vf/kdre rhozrk 2v0 vko`fÙk ds lehi gksxhA (B) fofdj.k dh vf/kdre rhozrk v0/ 2 vko`fÙk ds lehi gksxhA (C*) dqy mRlftZr ÅtkZ 16 ds xq.kkad ls c<+ tk;sxhA (D) dqy mRlftZr ÅtkZ 8 ds xq.kkad ls c<+ tk;sxhA
6.
A solid sphere and a hollow sphere of the same material and of equal radii are heated to the same temperature.
leku inkFkZ ,oa cjkcj f=kT;k ds ,d Bksl xksys rFkk ,d [kks[kys xksys dks leku rki rd xeZ djrs gS& (A*) Both will emit equal amount of radiation per unit time in the beginning
izkjEHk esa nksuksa ,dkad le; esa leku ek=kk dk fofdj.k mRlftZr djsaxsA (B*) Both will absorb equal amount of radiation from the surrounding in the beginning
izkjEHk esa nksuksa leku ek=kk dk fofdj.k okrkoj.k ls vo'kksf"kr djsxa As (C) The initial rate of cooling will be the same for the two spheres
nksuksa xksyks ds fy, izkjfEHkd 'khryu nj leku gksxhA (D) The two spheres will have equal temperatures at any instant.
[Q.6/HCV-II/CH-28]
fdlh Hkh {k.k nksuksa xksyks dk rki leku gksxkA manishkumarphysics.in
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Chapter # 28
Heat Transfer
dQ
Sol.
Since for emitting dt = eAT 4 e dQ = eAT T s4 dt Therefore, Both with emit & absorb equal amount of radiation in the beginning.
And for absorption ;
dT eA 4 = T dt ms All the terms on RHS are same for both spheres except mass "m". ( Solid and hollow)
For cooling
Hence
dT is different for both. dt
Hence (C) and (D) are wrong.
WORKED OUT EXAMPLES 1.
Sol.
The lower surface of a slab if stone of face-area 3600 cm2 and thickness 10 cm is exposed to steam at 100ºC. A block of ice at 0ºC rests on the upper surface of the slab 4.8g of ice melts in one hour. Calculate the thermal conductively of the stone. Latent heat of fusion of ice = 3.36 × 105 J/kg. The amount of heat transferred through the slab to the ice in one hour is Q = (4.8 × 10–3kg) × (3.36 × 105 J/kg) = 4.8 × 336 J. Using the equation Q=
KA(1 2 )t , x
K(3600cm 2 )(100º C) (3600 s) 4.8 × 336 J = 10 cm or, 2.
Sol.
K = 1.24 × 10–3 W/m–ºC.
An ice box made of 1.5 cm thick styrofoam has dimensions 60 cm × 60 cm × 30 cm. It contains ice at 0ºC and is keot in a room at 40ºC. Find the rate at which the ice is melting. Latent heat of fusion of ice = 3.36 × 105J/kg. and thermal conductively of styrofoam = 0.04 W/m–ºC. The total surface area of the walls = 2(60 cm × 60 cm + 60 cm × 30 cm + 60 cm × 30 cm) = 1.44 m2. The thickness of the walls = 1.5 cm = 0.015 m. The rate of heat flow into the box is Q KA (1 2 ) t x
=
(0.04 W / mº C)(1.44 m2 )( 40º C) 154 W 0.015 m
The rate in which the ice melts is 154 W
= 3.
3.36 10 5 J / kg
= 0.46 g/s.
A colosed cubical box is made of perfectly insulating material and the only way for heat to enter or leave the box is through two solid cylindrical metal plugs, each of cross-sectional area 12 cm2 and length 8 cm fixed in the opposite walls of the box. The outer surface of one plug is kept at a temperature of 100ºC while the outer surface of the other plug is maintained at a temperature of 4ºC. The thermal conductivity of the material of the plug is 2.0 W/m— ºC. A source of energy generating 13W is enclosed inside the box assuming that it is the same at all points on the inner surface..
Sol.
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Chapter # 28 Heat Transfer The situation is shown in figure. Let the temperature inside the box be . The rate at which heat enters the box through the left plug is Q1 KA (1 ) t x The rate of heat generation in the box = 13 W. The rate at which heat flows out of the box through the right plug is Q 2 KA ( 2 ) t x In the steady state Q1 Q 2 13 W t t
or
KA KA (1 ) 13 W ( 2 ) x x
or,
2
or,
2 (13 W )x 1 2 2KA =
KA KA (1 2 ) 13 W x x
100 º C 4º C (13 W ) 0.08m 2 2 (2.0 W / m º C) (12 10 4 m 2 )
= 52ºC + 216.67ºC = 269ºC. 4.
A bar of copper of length 75 cm and a bar of steel of length 125 cm are joined together end. Both are of circular cross-section with diameter 2 cm. The free ends of the copper and the steel bars are mintained at 100ºC and 0ºC respectively. The curved surface of the bars are thermally insulated. What is the temperature of the copper-steel junction ? What is the amount of heat transmitted per unit time across the junction? Thermal conductivity of copper is 386 J/m-s-ºC and that of steel is 46ºJ/m-s-ºC.
Sol. The situtation is shown in figure. Let the temperature at the junction be (on Celsius scale). The same heat current passes through the copper and the steel rods. Thus , K steel A Q K cu A(100 º C ) = t 75cm 125cm
or,
K cu (100 º C ) K steel 75 125
or,
100 º C 75K steel 3 46 125 K cu = 5 386
or, The rate of heat flow is Q t
K steel A = 125 cm ( 46J / m s º C) ( 1cm 2 ) 93 º C 125cm = 1.07 J/s.
=
5.
Two parallel plates A and B are joined together to form a compound plate (figure). The thicknesses of the plates are 4.0 cm and 2.5 cm respectively and the area of cross-section is 100 cm2 for each plate. The thermal conductivities are KA = 200 W/m-ºC for the plate A and KB = 400 W/m-ºC for the plate B. The outer surface of the plate A is maintained at 100ºC and the outer surface of the plate B is maintained at 0ºC. Find (a) the rate of heat flow through any cross-section, (b) the temperature at the interface and (c) the equivolent thermal conductivity of the compound plate.
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Chapter # 28 Heat Transfer Sol. (a) Let the temperature of the interface be . The area of cross-section of each plate is A = 100 cm3 = 0.01 m2. The thickness are xA = 0.04 m and xB = 0.025 m. The thermal resistance of the plate A is
1 xA KA A
R1
and that of the plate B is
R2
1 xB KB A
The equvalent thermal resistance is
R R1 R 2
Thus,
1 x A xB A K A K B
.....................(i)
Q 1 2 t R
A(1 2 ) = x /K x /K A A B B =
(0.01m2 ) (100 º C) (0.04m) (200 W / m º C) (0.025m) / ( 400 W / mº C)
= 3810 W.
Q A( 2 ) (b) We have t x / K B B or,
3810 W =
(0.01m2 ) ( 0º C) (0.025 m) / ( 400 W / mº C)
or, = 24ºC (c) If K is the equivalent thermal conductivity of the compound plate, its thermal resistance is R=
1 x A xB A K
Comparing with (i),
x A xB x x A B K K A KB or,
6.
Sol.
x A xB K = x / x /K A A B B = 248 W/m-ºC
A room has a 4 m × 4 m × 10 cm concrete roof (K = 1.26 W/m-ºC). At some instant, the temperature outside is 46ºC and that inside is 32ºC. (a) Negalecting convection, calculate the amount of heat flowing per second into the room through the roof. (b) Bricks ( K = 0.65 W/m-ºC) of thickness 7.5 cm are laid down on the roof. Calculate the new rate of heat flow under the same temperature conditions. The are of the roof = 4 m × 4 m = 16 m2. The thickness x = 10 cm = 0.10 m. (a) The thermal resistance of the roof is
1 x 1 0.10m R1 = K A 1.26 W / mº C 16m2 = 4.96 × 10-3 ºC/W 1 2 Q 46 0 C 32 0 C = R = t 1 4.96 10 3 c / w (b) The thermal resistance of the brick layer is Heat current is
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Chapter # 28
Heat Transfer R2
1 x 1 7.5 10 2 m K A 0.65 W / m º C 16 m 2
= 7.2 × 10-3 ºC/W. The equivalent thermal resistance is R = R1 + R2 = (4.96 + 7.2) × 10-3 ºC/W = 1.216 × 10-2 ºC/W. The heat current is Q 1 2 46 º C 32º C t R 1.216 10 3 º C / W
7.
= 1152 W An electric heater is used in a room of total wall area 137 m2 to maintain a temperature of 20ºC inside it, when the outside temperature is -10ºC. The walls have three different layers of materials. The innermost layer is of wood of thickness 2.5 cm, the middle layer is of cement of thickness 1.0 cm and the outermost layer is of brick of thickness 25.0 cm. Find the power of the electric heater. Assume that there is no heat loss through the floor and the ceiling. The thermal conductivities of wood, cement and ‘brick are 0.125W/m-ºC, 1.5 W/m-ºC and 1.0 W/m-ºC respectively.
Sol.
The situation is shown in figure. The thermal resistance of the wood, the cement and the brick layers are Rw
and
=
1 x K A
=
1 2.5 10 2 0.125 W / mº C 137m 2
RC
1 1.0 10 2 m = 1.5 Wm º C 137 m 2
RB
1 25.0 10 2 m = 1.0 W / m º C 137 m 2
=
0.20 ºC/ W 137
=
0.0067 ºC/ W 137
0.25 º C / W. 137 As the layers are connected in series, the equivalent thermal resistance is R = Rw + Rc + RB
=
0.20 0.0067 0.25 ºC/ W 137 = 3.33 × 10-3 ºC/W. The heat current is
=
1 2 20 º C ( 10 º C) = = 9000 W.. R 3.33 10 3 º C / W The heater must supply 9000 W to compenstate the outflow of heat. i
8.
=
Three rods of material x and three of material y are connected as shown in figure. All the rods are indentical in length and cross-sectional area. If the end A is maintained at 60ºC and the junction E at 10ºC, calculate the temperature of the junction B. The thermal conductivity of x is 800 W/m-ºC and that of y is 400 W/m-ºC.
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Chapter # 28 Heat Transfer Sol. It is clear from the symmetry of the figure that the points C and D are equivalent in all respect and hence, they are at the same temperature, say . No heat will flow through the rod CD. We can, therefore, neglect this rod in further analysis. Let and A be the length and the area of cross-section of each rod. The thermal resistance of AB BC and BD are equal. Each has a value
R1
1 l Kx A
........(i)
Similarly, thermal resistances of CE and DE are equal, each having a value
1 R2 = K A Y
........(ii)
As the rod CD has no effect, we can say that the rods BC and CE are joined in series. Their equivalent thermal resistance is R3 = RBC + RCE = R1 + R2 Also, the rods BD and DE together have an equivalent thermal resistance R4 = RBD + RDE = R1 + R2. The resistances R3 and R4 are joined in parallel and hence their equivalent thermal resistance is given by
R 3 R1 R 2 2 2 This resistance R5 is connected in series with AB. Thus the total arrangement is equivalent to a thermal resistance 1 1 1 2 R5 R3 R 4 R3
or,
R5
R1 R 2 3R1 R 2 2 2 Figure shows the sucessive steps in this reduction. R R AB R 5 R1
The heat current through A is
2( A E ) A E = 3R R R 1 2 This current passes through the rod AB. We have i=
i= or,
A n R AB
A – B = (RAB) i
2( A E ) = R1 3R R . 1 2 Putting from (i) and (ii), A –B = or, 9.
2k y ( A E ) K x 3K y
=
2 400 × 50ºC = 20ºC 800 3 400
B = A – 20ºC = 40ºC.
A rod CD of thermal resistance 5.0 K/W is joined at the middle of an identical rod AB as shown in figure. The ends A, B and D are maintained at 100ºC, 0ºC and 25ºC respectively. Find the heat current in CD.
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Chapter # 28 Heat Transfer Sol. The thermal resistance of AC is equal to that of CD and is equal to 2.5 K/W. Suppose, the temperature at C is . The heat currents through AC, CB and CD are Q1 100 º C = . t 2.5K / W Q 2 0º C t 2.5K / W Q 3 25 º C . t 5.0K / W
and We also have
Q 3 Q1 Q 2 = + t t t
or, or, or, Thus,
100 º C 0º C 25 º C = + 2 .5 2.5 5 225ºC = 5 = 45ºC
20 K Q 3 45 º C 25 º C = = 5.0 K / W = 4.0 W t 5.0K / W
10.
Two thin metallic spherical shells of radii r1 and r2 (r1 < r2) are placed with their centres coinciding. A material of thermal conductively K is filled in the space between the shells. The inner shell is maintained at temperature 1 and the outer shell at temperature 2 (1 < 2). Calculate the rate at which heat flows radially through the material.
Sol.
Let us draw two spherical shells of radii x and x + dx concentric with the given system. Let the temperatures at these shells be and + d respectively. The amount of heat of heat flowing radially inward through the material between x and x + dx is
Q K 4x 2 d t dx Thus, 2
r2
1
1
Q dx K 4 d t x 2 r
or,
or, 11.
Q 1 1 K 4(2 – 1) = t r r 1 2 Q 4Kr1r2 ( 2 1 ) t r2 r1
On a cold winter day, the atmospheric temperature is – (on Celsius scale) which is below 0ºC. A cylindrical drum of height h made of a bad conductor is completely filled with water at 0ºC and is kept outside without any lid. Calculate the time taken for the whole mass of water to freeze. Thermal conductively of ice is K and its latent heat of fusion is L. Neglect expansion of water on freezing.
Sol.
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Page # 10
Chapter # 28 Heat Transfer Suppose, the ice starts forming at time t = 0 and a thickness x is formed at time t. The amount of heat flown from the water to the surrounding in the time interval t to t + dt is
KA dt x The mass of the ice formed due to the loss of this amount of heat is Q
Q KA dt. L xL The thickness dx of ice formed in time dt is dm=
dx
dm K dt Ap xL
L x dx . K Thus, the time T taken for the whole mass of water to freeze is given by
or,
dt =
T
0
h
L dt xdx K
0
T=
or,
Lh 2 2K
12.
Figure shows a large tank of water at a constant temperature 0 and a small vessel containing a mass m of water at an initial temperature 1(<0). A metal rod of length L, area of cross-section A and thermal conductivity K connects the two vessels. Find the time taken for the temperature of the water in the smaller vessel to become 2(1 < 2 < 0). Specific heat capacity of water is s and all other heat capacities are negligible.
Sol.
Suppose, the temperature of the water in the smaller vessel is at time t. In the next time interval dt, a heat Q is transferred to it where
KA ( 0 ) dt . ...........(i) L This heat increases the temperature of the water of mass m to + d where Q = ma d ...........(ii) From (i) and (ii), Q =
KA (0 – ) dt = ms d L or,
dt =
d Lms KA 0
T
or,
dt
0
Lms KA
2
1
d 0
where T is the time required for the temperature of the water to become 2. Thus, 13.
Sol.
T
1 Lms In 0 KA 0 2
One mole of an ideal monatomic gas is kept in a rigid vessel. The vessel is kept inside a steam chamber whose temperature is 97ºC, Initially, the temperature of the gas is 5.0ºC. The walls of the vessel have an inner surface of area 800 cm2 and thickness 1.0 cm. If the temperature of the gas increases to 9.0ºC in 5.0 seconds, find the thermal conductivity of the material of the walls. The initial temperature difference is 97ºC – 5ºC = 92ºC and at 5.0 s the temperature difference becomes 97ºC – 9ºC = 88ºC. As the change in the temperature difference is small, we work with the average temperature difference
manishkumarphysics.in
Page # 11
Chapter # 28
Heat Transfer
92º C 88 º C 90 º C 90K . 2 The rise in the temperature of the gas is 9.0ºC – 5.0ºC = 4ºC = 4K. The heat supplied to the gas in 5.0 s is Q = nCv T J 3 ( 4k ) = (1 mole) × 8.3 mol K 2 = 49.8 J. If the thermal conductivity is K, 49.8J
or 14.
Sol.
K(800 10 4 m 2 ) (90K ) 1.0 10 2 m K
× 5.0 s
49.8 J 0.014 J / m s K . 3600m s K
A monatomic ideal gas is contained in a rigid container of volume V with walls of total inner surface area A, thickness x and thermal conducticity K. The gas is at an initial temperature T0 and pressure p0. Find the pressure of the gas as a function of time if the temperatureof the surrounding air is T0. All temperatures are in absolute scale. As the volume of the gas is constant, a heat Q given to the gas increases its temperature by T = Q/Cv. Also, for a monatomic gas, C v
3 R. If the temperature of the gas at time t is T, the heat 2
current into the gas is Q KA(Ts T ) = t x
or,
T 2KA ( Ts T ) t 3 xR T
or,
T0
dT Ts T
l
2KA
3xR dt 0
or,
Ts T0 2KA In T T 3 xR t s
or,
Ts – T = (Ts – T0) e
or, T = Ts – (Ts – T0) e As the volume remains constant,
2KA t 3 xR
2KA t 3 xR
p p0 T T0 or,
P
p0 T T0
p 0 T0 15.
Sol.
2KA t Ts (Ts T0 ) e 3 xR .
Consider a cubical vessel of edge a having a small hole is one of its walls. The total thermal resistance of the walls is r. At time t = 0, it contains air at atmospheric pressure pa and temperature T0. The temperature of the surrounding air is Ta(>T0). Find the amount of the gas (in moles) in the vessel at time t. Take Cv of air to be 5 R/2. As the gas can leak out of hole, the pressure inside the vessel will be equal to the atmospheric pressure pa. Let n be the amount of the gas (moles) in the vessel at time t. Suppose an amount Q of heat is given to the manishkumarphysics.in
Page # 12
Chapter # 28 Heat Transfer gas in time dt. Its temperature increases by dT where Q = nCp dT If the temperature of the gas is T at time t, we have Q Ta T dt r or, (Cpr)n dT = (Ta – T) dt. We have, paa3 = nRT or, ndT + Tdn = 0 or, n dT = – T dn.
T
Also,
...........(i)
...........(ii) 3
p 0a nR
...........(iii)
Using (ii) and (iii) in (i),
Cprp 0 a 3 nR
p 0a 3 dn = Ta nR dt
dn p a nR Ta a nR 3
or,
where n0
dt Cprp a a 3
p aa 3 is the initial amount of the gas in the vessel. Thus, RT0
nRTa p a a 3 t 1 In 3 RTa n0RTa p a a Cprp a a 3
or,
nRTa – paa = (n0RTa – paa ) e 3
3
n0
Writing
RTa Cprpaa 3
t
.
p aa 3 7R and Cp C v R RT0 2
2Ta 7rpaa 3 t p a a 3 T0 n 1 1e . RTa T0
16.
Sol.
17. Sol.
A blackbody of surface area 1 cm2 is placed inside an enclosure. The enclosure has a constant temperature 27ºC and the blackbody is maintained at 327ºC by heating it electrically. What electric power is needed to maintain the temperature? = 6.0 × 10–8 W/m2 –K4. The area of the blackbody is A = 10–4 m2, its temperature is T1 = 327ºC = 600 K and the temperature of the enclosure is T2 = 27ºC = 300 K. The blackbody emits radiation at the rate of AT14. The radiation falls on it (and gets absorbed) at the rate of AT24. The net rate of loss of energy is A(T14 – T24). The heater must supply this much of power. Thus, the power needed is A(T14 – T24) = (10–4 m2) (6.0 × 10–8 W/m2 – K4) [(600 K)4 – (300 K)4] = 0.73 W. An electric heater emits 1000W of thermal radiation. The coil has a surface area of 0.020 m2. Assuming that the coil radiates like a blackbody, find its temperature. = 6.00 × 10–8 W/m2 – K4. Let the temperature of the coil be T. The coil will emit readiation at a rate AT4. Thus 1000 W = (0.020 m2) × (6.0 × 10–8 W/m2 – K4) × T4 or, or,
18.
T4
1000
0.020 6.00 10 8 = 8.33 × 1011 K4 T = 955 K.
K4
The earth receives solar radiation at a rate of 8.2 J/cm2 – minute. Assuming that the sun radiates like a blackbody, calculate the surface temperature of the sun. The angle subtended by the sun on the earth is 0.53º and the Stefan constant = 5.67 × 10–8 W/m2 – K4. manishkumarphysics.in
Page # 13
Chapter # 28
Heat Transfer
Sol. Let the diameter of the sun be D and its distance from the earth be R. From the question D 0.53 R 180 = 9.25 × 10–3. The radiation emitted by the surface of the sun per unit time is
.........(i)
2
D 4 T4 = D2 T4. 2 At distance R, this radiation falls on an area 4R2 in unit time. The radiation received at the earth’s surface per unit time per unit area is, therefore. 2
D 2T 4
4R 2
T 4 D . 4 R
2
Thus,
T 4 D = 8.2 J/cm2 - minute 4 R
or,
W 1 8 T4 × (9.25 × 10–3)2 × 5.67 10 2 4 m K 4
10 T = 5794 K = 5800 K.
or, 19. Sol.
8.2 4
60m 2
The temperature of a body falls from 40ºC to 36ºC in 5 minutes when placed in a surrounding of constant temperature 16ºC. Find the time taken for the temperature of the body to become 32ºC As the temperature differences are small, we can use Newton’s law of cooling. d k ( 0 ) dt
d kdt 0
or,
........(i)
where k is a constant, is the temperature of the body at time t and 0 = 16ºC is the temperature of the surrounding. We have.
30 º C
40 º C
or,
d = –k (5 min) In
36 º C 16º C = – k (5 min) 40º C 16 º C
In(5 / 6) 5 min If t be the time required for the temperature to fall from 36ºC to 32ºC then by (i), k
or,
32 º C
36 º C
or,
or,
In
d kt 0
32º C 16 º C In(5 / 6)t 36 º C 16 º C 5 min
t=
In( 4 / 5) 5 min In(5 / 6)
= 6.1 min
manishkumarphysics.in
Page # 14
Chapter # 28 Alternative method
Heat Transfer
The mean temperature of the body as it cools from 40ºC to 36ºC is
decrease of temperature is
40 º C 36 º C = 38ºC. The rate of 2
40 º C 36 º C = 0.80ºC/min. 5 min
Newton’s law of cooling is
or,
d = –k( – 0) dt –0.8ºC/min = –k(38ºC – 16ºC) = – k(22ºC)
0.8 min 1 22 Let the time taken for the temperature to become 32ºC be t. During this period, k
or,
d 36 º C 32º C 4º C . dt t t
The mean temperature is
36 º C 32º C 34 º C 2
Now, d = –k ( – 0) dt
or,
4º C 0 .8 × (34ºC – 16ºC)/min t 22
or, 20.
t
22 4 min = 6.1 min 0.8 18
A hot body placed in air is cooled down according to Newton’s law of cooling, the rate of decrease of temperature being k times the temperature difference from the surrounding. Starting from t = 0, find the time in which the body will lose half the maximum heat it can lose.
,d xeZ oLrq tks gok esa j[kh gqbZ gS] dks U;wVu ds 'khrye ds fu;e ls B.Mk fd;k tkrk gSA rkieku ds fxjus dh nj okrkoj.k ls rkieku vUrj dk k xquk gSA t = 0 ls 'kq: djrs gq,] ml le; dk eku crkb, tc ;g oLrq vf/kdre Å"ek dk vk/ kh Å"ek dk ºkl dj pqdh gSA HCV_Ch-28_WOE_20 Sol.
We have, d = – k( – 0) dt where 0 is the temperature of the surrounding and is the temperature of the body at time t. Suppose = 1 at t = 0. Then,
1
or,
In
d 0 = –k
t
dt 0
0 1 0 = –kt
or, – 0 = (1 – 0) e–kt. ........(i) The body continues to lose heat till its temperature becomes equal to that of the surrounding. The loss of heat in this entire period is Qm = ms(1 – 0). This is the maximum heat the body can lose. If the body loses half this heat, the decrease in its temperature will be, Q m 1 0 2ms 2 If the body loses this heat in time t1, the temperature at t1 will be
manishkumarphysics.in
Page # 15
Chapter # 28
Heat Transfer
1 0 1 0 2 2 Putting these valuesof time andtemperature in (i), 1
1 0 – 0 = (1 – 0) e kt1 2 or,
e kt1
or,
t1
1 2
In 2 . k
EXERCISE 1.
Ans.
A uniform slab of dimension 10cm × 10cm × 1cm is kept between two heat reservoirs at temperatures 10ºC and 90ºC. The larger surface areas touch the reservoirs. The thermal conductivity of the material is 0.80 W/ m–ºC. Find the amount of heat flowing through the slab per minute. 10 lseh × 10 lseh × 1 lseh foekvksa okyh ,d le:i ifV~Vdk dks nks Å"ek /kkjdksa ds chp j[kk x;k gSA ftuds rki 10ºC rFkk 90ºC gSA vf/kd {ks=kQy okys Qyd Å"ek/kkjdksa ds lEidZ esa gSA inkFkZ dh Å"ek pkydrk 0.80 okWV@eh–ºC gSA ifV~Vdk ls izfr fefuV izokfgr gksus okyh Å"ek dh ek=kk Kkr dhft;sA 3840 J
2.
A liquid-nitrogen container is made of a 1 cm thick thermocoal sheet having thermal conductivity 0.025 J/m– s–ºC. Liquid nitrogen at 80 K is kept in it. A total area of 0.80m2 is in contact with the liquid nitrogen. The atmospheric temperature is 300 K. Calculate the rate of heat flow from the atmosphere to the liquid nitrogen. rjy ukbVªkt s u ls Hkjk ,d ik=k 1 cm eksVkbZ o Å"eh; pkydrk 0.025 J/m–s–ºC dh Fkeksd Z ksy dh 'khV ls cuk gSA blesa 2 rjy ukbVªkstu dks 80 K rkieku ij j[kk tkrk gSA bles dqy 0.80m {ks=kQy rjy ukbVªkt s u ds lEidZ esa gSA ok;qe.My dk rkieku 300 K gSA ok;qe.My ls rjy ukbVªkstu dh vksj Å"ek ds cgus dh nj dh x.kuk dhft,A
Ans.
440 W
3.
The normal body-temperature of a person is 97ºF. Calculate the rate at which heat is flowing out of his body through the clothes assuming the following values. Room temperature = 47ºF, surface of the body under clothes = 1.6 m2, conducitivity of the cloth = 0.04 J/m–s–ºC, thickness of the cloth = 0.5 cm. lkekU; ekuo 'kjhj dk rki 97ºF gksrk gSA diM+kas ls gksdj mlds 'kjhj ls ckgj dh vksj izokfgr Å"ek dh nj dh x.kuk vxzkafdr ekuksa ds v/kkj ij dhft;s] dejs dk rki = 47ºF, diM+ksa ds uhps 'kjhj dh lrg dk {ks=kQy = 1.6 eh2, diM+ksa dh pkydrk = 0.04 twy/eh–s–ºC, diM+ksa dh eksVkbZ = 0.5 lseh Ans. 356 J/s
4.
Water is boiled in a container having a bottom of surface area 25cm2, thickness 1.0 mm and thermal conductivity 50 W/mºC, 100 g of water is converted into steam per minute in the steady state after the boiling starts. Assuming that no heat is lost to the atmosphere, calculate the temperature of the lower surface of the bottom. Latent heat of vaporization of water = 2.26 × 106 J/kg. ,d ik=k esa ikuh mckyk tk jgk gSA ftlds iSna s dh lrg dk {ks=kQy 25 lseh2, eksVkbZ 1.0 feeh rFkk Å"ek pkydrk 50 okWV@eh– ºC, 100 xzke ikuh izfr fefuV Hkki esa ifjofrZr djuk gSA ;g ekurs gq, fd okrkoj.k esa dksbZ Å"ek gkfu ugha gksrh
HCV_Ch-28_Ex._2
gS] iSans dh fupyh lrg ds rki dh x.kuk dhft;sA ikuh ds ok"iu dh xqIr Å"ek = 2.26 × 106 twy/fdxzk. Ans.
130°C
5.
One end of a steel rod (K = 46 J/m–s–ºC) of length 1.0 m is kept in ice at 0ºC and the other end is kept in boiling water at 100ºC. The area of cross–section of the rod is 0.04cm2. Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice = 3.36 × 105 J/kg. 1.0 eh yEch LVhy dh NM+ (K = 46 twy/eh–ls–ºC) dk ,d fljk 0ºC ds cQZ esa rFkk nwljk fljk 100ºC ds mcyrs gq, ty esa j[kk x;k gSA NM+ dk vuqiLz Fk dkV 0.04 lseh2 gSA ;g ekurs gq, fd ok;qe.My dks dksbZ Å"ek gkfu ugha gksrh gS] izfr lsd.M
fi?kyus okys cQZ dh ek=kk Kkr dhft;sA cQZ ds xyu dh xqIr Å"ek = 3.36 × 105 twy/fdxzk. Ans.
5.5 × 10–2 g
manishkumarphysics.in
Page # 16
Chapter # 28 Heat Transfer 6. An icebox almost completely filled with ice at 0ºC is dipped into a large volume of water at 20ºC. The box has walls of surface area 2400cm2, thickness 2.0 mm and thermal conductivity 0.06 W/m–ºC. Calculate the rate at which the ice melts in the box. Latent heat of fusion of ice = 3.4 × 105 J/kg. 0ºC rki okyh cQZ lsyxHkx iwjk Hkjk gqvk ,d vkbl&ckWDl] 20ºC rki okys ikuh ds ,d cgqr cM+s vk;ru esa iwjk Mwck fn;k x;k gSA ckWDl dh nhokjksa dk i`"B&{ks=kQy 2400 lseh2, eksVkbZ 2.0 feeh rFkk Å"ek pkydrk 0.06 okWV/eh–ºC gSA ckWDl esa cQZ ds xyu dh xqIr Å"ek = 3.4 × 105 twy@fdxzkA Ans. 1.5 kg/h 7.
Ans. 8.
A pitcher with 1mm thick porous walls contains 10 kg of water. Water comes to its outer surface and evaporates at the rate of 0.1 g/s. The surface area of the pitcher (one side) = 200 cm2. The room temperature = 42ºC, latent heat of vaporization = 2.27 × 106J/kg, and the thermal conductivity of the porous walls = 0.80 J/m–s–ºC. Calculate the temperature of water in the pitcher when it attains a constant value. 1 feeh eksVkbZ ,oa ljU/kz nhokjksa okys eVds esa 10 fdxzk ikuh gSA ikuh bldh ckgj okyh lrg ij vkdj 0.1 xzke@ls dh nj ls okf"ir gksrk jgrk gSA ?kM+s dh lrg dk {ks=kQy ¼,d vksj dk½ = 200 lseh2 gSA dejs dk rki = 42ºC, ok"iu dh xqIr Å"ek = 2.27 × 106 twy/fdxzk rFkk ljU/kz nhokjksa dh Å"ek pkydrk = 0.80 twy/eh–ls–ºC gSA ?kM+s esa ikuh ds rki dh x.kuk dhft;s] tc bldk eku fLFkj gks tkrk gSA 28°C A steel frame (K = 45 W/m–ºC) of total length 60cm and cross-sectional area 0.20 cm2, forms three sides of a square. The free ends are maintained at 20ºC and 40ºC. Find the rate of heat flow through a cross-section of the frame. ,d LVhy&Ýse (K = 45 okWV/eh–ºC) ftldh dqy yEckbZ 60 lseh rFkk vuqiLz Fk dkV dk {ks=kQy 0.20 lseh2, gS] ,d o"kZ dh rhu Hkqtk,¡ cukrh gSA eqDr fljs 20ºC rFkk 40ºC rki ij j[ks x;s gSAa Ýse dh fdlh vuqiLz Fk dkV ls izokfgr Å"ek dh
nj Kkr dhft;sA Ans.
0.03 W
9.
Water at 50ºC is filled in a closed cylindrical vessel of height 10 cm and cross-sectional area 10cm2. The walls of the vessel are adiabatic but the flat parts are made of 1 mm thick aluminium (K = 200 J/m–s–ºC). Assume that the outside temperature is 20ºC. The density of water is 1000 kg/m3, and the specific heat capacity of water = 4200 J/kg 0C. Estimate the time taken for the temperature to fall by 1.0ºC. Make any simplifying assumptions you need but specific them. 10 lseh Å¡ps ,oa 10 lseh2 vuqiLz Fk dkV {ks=kQy okys ,d can csyukdkj ik=k esa 50ºC ij ikuh Hkjk gqvk gSA ik=k dh nhokjsa :nks"e gS fdUrq blds lery Hkkx 1 feeh eksVh ,Y;wfefu;e (K = 200 twy@eh–ls–ºC) ls cus gq, gSAa ckgj dk rki 20ºC eku yhft;sA ikuh dk ?kuRo 1000 fdxzk@eh3, rFkk ikuh dh fof'k"V Å"ek = 4200 twy@fdxzk 0C gSA rki esa 1.0ºC deh
gksus esa yxus okys le; dk vuqeku yxkb;sA ljyhdj.k ds fy;s vki dqN /kkj.kk,¡ cuk ldrs gS]a fdUrq mudks O;Dr dhft;sA Ans.
0.035 s
10.
The left end of a copper rod (length = 20 cm, area of cross-section = 0.20cm2) is maintained at 20ºC and the right end is maintained at 80ºC. Neglecting any loss of heat through radiation find (a) point 11 cm from the left end and (b) the heat current through the rod. Thermal conductivity of copper = 385 W/m–ºC. rkacs dh ,d NM+ (yEckbZ = 20 lseh, vuqiLz Fk dkVk dk {ks=kQy = 0.20 lseh2) dk cka;k fljk 20ºC rki ij j[kk x;k gS rFkk nk;ka fljk 80ºC rki ij j[kk x;k gSA fofdj.k }kjk Å"ek gkfu ux.; ekurs gq, Kkr dhft;s : (a) ck;sa fljs ls 11 lseh nwj fLFkr fcUnq dk rki rFkk (b) NM+ ls izokfgr Å"ek&/kkjkA rkacs dh Å"ek pkydrk = 385 okWV/eh–ºC. (a) 53°C (b) 2.31 J/s
Ans. 11.
The ends of a metre stick are maintained at 100ºC and 0ºC. One end of a rod is maintained at 25ºC. Where should its other end be touched on the metre stick so that there is no heat current in the rod in steady state? ,d ehVj NM+ ds fljksa ds rki 100ºC rFkk 0ºC j[ks x;s gSaA ,d NM+ dk ,d fljk 25ºC rki ij gSA bldk nwljk fljk
ehVj&NM+ ij dgk¡ ij Nqvk fn;k tk;s fd LFkk;h fLFkfr esa NM+ ls dksbZ Å"ek&/kkjk izokfgr ugha gks\ Ans.
25 cm from the cold end.
12.
A cubical box of volume 216 cm3 is made up of 0.1 cm thick wood. The inside is heated electrically by a 100 W heater. It is found that the temperature difference between the inside and the outside surface is 5ºC in steady state. Assuming that the entire electrical energy spent appears as heat, find the thermal conductivity of the material of the box. 216 lseh3 vk;ru okyk ,d ?kukdkj ckWDl 0.1 lseh eksVh ydM+h ls cuk gqvk gSA ;g vUnj dh vksj 100 okWV ds ghVj ls xeZ fd;k tk jgk gSA LFkk;h fLFkfr esa vkarfjd rFkk cká lrgksa ds e/; 5ºC rkikarj izfs {kr fd;k x;k gSA eku yhft;s fd
O;kf;r gksus okyh lEiw.kZ fo|qr&ÅtkZ] Å"ek ds :i esa izdV gksrh gS] ckWDl ds inkFkZ dh Å"ek pkydrk Kkr dhft;sA Ans.
0.92 W/m–°C manishkumarphysics.in
Page # 17
Chapter # 28 Heat Transfer 13. Figure shows water in a container having 2.00 mm thick walls made of a material of thermal conductivity 0.50 W/m–ºC. The container is kept in a melting-ice bath at 0ºC. The total surface area in contact with water is 0.05 m2. A wheel is clampled inside the water and is coupled to a block of mass M as shown in the figure. As the block goes down, the wheel rotates. It is found that after some time a steady state is reached in which the block goes down with a constant speed of 10cm/s and the temperature of the water remains constant at 1.0C. Find the mass M of the block. Assume that the heat flows out of the water only through the walls in constant. Take g = 10 m/s2. fp=k esa ,d ik=k esa ikuh iznf'kZr fd;k x;k gSA ftldh 2.00 feeh eksVh nhokjsa 0.50 okWV/eh–ºC Å"ek pkydrk okys inkFkZ dh cuh gqbZ gSA ik=k dks 0ºC rki ij fi?kyrs gq, cQZ dh ukn esa j[kk x;k gSA ikuh ds lkFk lEidZ dk dqy {ks=kQy 0.05 m2 gSA ikuh esa,d ifg;k yx;k x;k gSA tks ,d Mksjh dh lgk;rk ls fp=kkuqlkj M nzO;eku okys CykWd ls tqM+k gqvk gSA
tSl&s tSls CykWd uhps dh vksj vkrk gS] ifg;k ?kwerk gSA ;g izfs {kr fd;k x;k gS fd dqN le; i'pkr~ LFkk;h LFkfr izkIr gks tkrh gSA ftlesa CykWd 10 lseh@ls fu;r pky ls uhps dh vksj tkrk gS rFkk ikuh dk rki 1.0° C fu;r cuk jgrk gSA CykWd dk nzO;eku M Kkr dhft;sA eku yhft;s fd ikuh ls ckgj dh vksj izokfgr gksus okyh Å"ek dsoy lEidZ esa vkus okyh nhokjksa ls gh gksrh gSA (g = 10 m/s2)
Ans.
12.5 kg
14.
On a winter day when the atmospheric temperature drops to –10ºC, ice forms on the surface of a lake. (a) Calculate the rate of increase of thickness of the ice when 10 cm of ice is already formed. (b) Calculate the total time taken in forming 10cm of ice. Assume that the temperature of the entire water reaches 0ºC before the ice starts forming. Density of water = 1000kg/m3, latent heat of fusion of ice = 3.36 × 105 J/kg and thermal conducitivty of ice = 1.7 W/m–ºC. Neglect the expansion of water on freezing. lnhZ ds fnuksa es]a tc ok;qe.My dk rki fxjdj –10ºC gks tkrk gS] >hy dh lrg ij cQZ te tkrh gSA (a) tc 10 lseh cQZ igys ls te pqdh gksrh gS] cQZ dh eksVkbZ c<+us dh nj dh x.kuk dhft;sA (b) 10 lseh cQZ teus esa yxus okys dqy le; dh x.kuk dhft;sA eku yhft;s fd cQZ teuk 'kq: gksus ls iwoZ lEiw.kZ ikuh dk rki 0°C gks pqdk FkkA ikuh dh xqIr Å"ek = 3.36 × 105 twy@fdxzk rFkk cQZ dh Å"ek pkydrk = 1.7 okWV@eh °C A teus ij cQZ dk izlkj ux.; eku yhft;sA (a) 5.0 × 10–7 m/s (b) 27.5 hours
Ans. 15.
Consider the situation of the previous problem. Assume that the temperatures of the water at the bottom of the lake remains constant at 4ºC as the ice forms on the surface(the heat required to maintain the temperature of the bottom layer may come from the bed of the lake). The depth of the lake is 1.0m. Show that the thickness of the ice formed attains a steady state maximum value. Find this value. The thermal conductivity of water = 0.50 W/m–ºC. Take other relevant data from the previous problem.
fiNys iz'u esa of.kZr fLFkfr ij fopkj dhft;sA eku yhft;s fd tc lrg ij cQZ teuk izkjEHk gksrh gSA >hy ds iSna as esa ikuh dk rki 4ºC fu;r cuk jgrk gSA ¼iSns ds lehi lrg dk rki cuk;s j[kus ds fy;s vko';d Å"ek >hy dh ryh ls vk ldrh gS½ >hy dh xgjkbZ 1.0 eh- gSA O;Dr dhft;s fd cuus okyh cQZ dh eksVkbZ ,d vf/kdre lkE;koLFkk eku izkIr djrh gSA ;g eku Kkr dhft;sA ikuh dh Å"ek pkydrk = 0.50 okWV@eh–ºC A vU; vko';d vkadM+s fiNys iz'u ls ys yhft;sA Ans.
89 cm
16.
Three rods of lengths 20cm each and area of cross-section 1 cm2 are joined to form a triangle ABC. The conductivities of the rods are KAB = 50 J/m–s–ºC, KBC = 200 J/m–s–°C and KAC = 400 J/m–s–°C. The junction A, B and C are maintained at 40°C, 80°C and 80°C respectively. Find the rate of heat flowing through the rods AB, AC and BC. rhu NM+as bl izdkj tksMh+ x;h gS fd ;s ,d f=kHkqt ABC cukrh gSA izR;sd NM+ dh yEckbZ 20lseh ,oa vuqiLz Fk dkVk dk {ks=kQy 1 lseh2 gSA NM+ksa dh pkydrk,¡ KAB = 50 twy@eh–s–ºC, KBC = 200 twy@eh–ls–°C vkSj KAC = 400 twy/eh–ls–°C gSA laf/ k;ksa A, B rFkk C ds rki Øe'k% 40°C, 80°C rFkk 80°C gSA NM+ksa AB, AC rFkk BC ls Å"ek izokg dh nj Kkr dhft;sA 1 W, 8 W, zero
Ans.
manishkumarphysics.in
Page # 18
Chapter # 28 Heat Transfer 17. A semicircular rod is joined at its end to a straight rod of the same material and the the same cross-sectional area. The straight rod forms a diameter of the other rod. The junctions are maintained at different temperatures. Find the ratio of the heat transferred through a cross-section of the semicircular rod to the heat transferred through a cross-section of the straight rod in a given time.
,d v)ZoÙ` kkdkj NM+ ds fljksa ij leku inkFkZ ,oa leku vuqiLz Fk dkV {ks=kQy okyh ,d lh/kh NM+ ls tksM+h x;h gSA lh/ kh NM+ nwljh NM+ dk O;kl cukrh gSA laf/k;ksa dks fHkUu&fHkUu rkiksa ij j[kk x;k gSA fdlh fn;s x;s le; esa v)ZoÙ` kkdkj NM+ ds vuqiLz Fk dkVj ls LFkkukarfjr Å"ek rFkk lh/kh NM+ ds vuqiLz Fk dkV ls LFkkukarfjr Å"ek rFkk lh/kh NM+ ds vuqiLz Fk dkV ls LFkkukarfjr Å"ek dk vuqikr Kkr dhft;sA Ans.
2:
18.
A metal rod of cross-sectional area 1.0 cm2 is being heated at one end. At one time, the temperature gradient is 5.0°C/cm at cross-section A and is 2.5°C/cm at cross-section B. Calculate the rate at which the temperature is increasing in the part AB of the rod. The heat capacity of the part AB = 0.40 J/°C, thermal conductivity of the material of the rod = 200 W/m–°C. Neglect any loss of heat to the atmosphere. 1.0 lseh2 vuqiLz Fk dkV {ks=kQy okyh /kkrq dh ,d NM+ dks ,d fljs ij xeZ fd;k tk jgk gSA fdlh le; ij vuqiLz Fk dkV A ij rki izo.krk 5.0°C/lseh rFkk vuqiLz Fk dkV B ij 2.5°C/lseh gSA x.kuk dhft;s fd NM+ ds Hkkx AB dh Å"ek /kkfjrk = 0.40 twy/°C, A okrkoj.k dks gksus okyh Å"ek gkfu ux.; eku yhft;sA 12.5°C/s
Ans. 19.
Ans. 20.
Ans.
Steam at 120°C is continuously passed through a 50 cm long rubber tub of inner and outer radii 1.0 cm and 1.2 cm. The room temperature is 30°C. Calculate the rate of heat flow through the walls of the tube. Thermal conductivity of rubber = 0.15 J/m–s–°C. 120°C dh Hkki 50 lseh yEch jcj dh uyh ls fujUrj xqtj jgh gSA bldh vkarfjd ,oa cká f=kT;k,¡ Øe'k% 1.0 lseh vkSj 1.2 lseh gSa dejs dk rki 30°C gSA uyh dh nhokjksa ls Å"ek ds izokg dh nj dh x.kuk dhft;sA jcj dh Å"ek pkydrk = 0.15 twy@eh–ls–°C gSA 233 J/s A hole of radius r1 is made centrally in a uniform circular disc of thickness d and radius r2. The inner surface (a cylinder of length d and radius r1) is maintained at a temperature 1 and the outer surface (a cylinder of length d and radius r2) is maintained at a temperature 2(1 > 2). The thermal conductivity of the material of the disc is K. Calculate the heat flowing per unit time through the disc. r1 f=kT;k dk ,d fNnz] d eksVkbZ rFkk r2 f=kT;k dh ,d le:i pdrh ds dsUnz ij cuk;k x;k gSA vkarfjd lrg (d yEckbZ o r1 f=kT;k ds csyu dh) dk rki 1 j[kk tkrk gSA rFkk cká lrg (d yEckbZ o r2 f=kT;k ds csyu dh) dk rki 2j[kk tkrk gSA 2(1 > 2) A pdrh ds inkFkZ dh Å"ek pkydrk K gSA ,dkad le; esa pdrh ls gksus okyh Å"ek izokg dh x.kuk dhft;sA
2Kd(1 2 ) In (r2 / r1 )
21.
A hollow tube has a length , inner radius R1 and outer radius R2. The material has thermal conductivity K. Find the heat flowing through the walls of the tube if (a) the flat ends are maintained at temperatures T1 and T2 (T2 > T1) (b) the inside of the tube is maintained at temperature T1 and the outside is maintained at T2. ,d [kks[kyh uyh dh yEckbZ , vkarfjd f=kT;k R1 rFkk cká f=kT;k R2 gSA blds inkFkZ dh Å"ek pkydrk K gSA uyh dh nhokjksa ls Å"ek izokg Kkr dhft;s] ;fn (a) lery fljksa ds rki T1 rFkk T2(T2 > T1) fLFkj j[ks tkrs gSAa (b) uyh dk vkarfjd Hkkx T1 rki ij rFkk cká Hkkx T2 rki ij j[kk tkrk gSA
Ans.
(a)
K(R 22 R12 ) (T2 T1 ) l
(b)
2Kl (T2 T1 ) In (R 2 / R1 )
22.
A composite slab is prepared by pasting two plates of thicknesses L1 and L2 and thermal conductivities K1 and K2. The slabs have equal cross-sectional area. Find the equivalent conductivity of the slab. L1 rFkk L2 eksVkbZ okyh ,oa K1 rFkk K2 Å"ek pkydrkvksa okyh nks ifV~Vdkvksa dks tksM+dj ,d la;D q r ifV~Vdk cuk;h tkrh
gSA ifV~Vdkvksa dk vuqizLFk dkVj ,d leku gSA ifV~Vdk dh rqY; pkydrk Kkr dhft;sA Ans. 23.
K 1K 2 (L1 L 2 ) L1L 2 L 2K 1 Figure shows a copper rod joined to a steel rod. The rods have equal length and equal cross-sectional area. The free end of the copper rod is kept at 0°C and that of the steel rod is kept at 100°C. Find the temperature at the junction of the rods. Conductivity of copper = 390 W/m–°C and that of steel = 46 W/m–°C. manishkumarphysics.in
Page # 19
Chapter # 28
Heat Transfer
fp=k esa rkacs dh NM+ ls tqMh+ gqbZ LVhy dh NM+ iznf'kZr gSA NM+kas dh yEckbZ;k¡ cjkcj gS rFkk vuqiLz Fk dkVj ,d leku gSA rkacs dh NM+ ds eqDr fljs dk rki 0°C j[kk x;k gS rFkk LVhy dh NM+ ds eqDr fljs dk rki 100°C gSA nksuksa NM+kas dh laf/ k dk rki Kkr dhft;sA rkacs dh pkydrk = 390 okWV@eh–°C rFkk LVhy dh pkydrk = 46 okWV/eh–°C gSA
Ans.
10.6°C
24.
An aluminium rod and a copper rod of equal length 1.0 m and cross-sectional area 1 cm2 are welded together as shown in figure. One end is kept at a temperature of 20°C and the other at 60°C. Calculate the amount of heat taken out per second from the hot end. Thermal conductivity of aluminium = 200 W/m–°C and of copper = 390 W/m–°C. ,d ,Y;wfefu;e dh NM+ rFkk ,d rkacs dh NM+ dh yEckbZ;k¡ ,d leku 1.0 eh- rFkk buds vuqiLz Fk dkV {ks=kQy ,d leku 1 lseh2 gS] vkil >ky nh x;h gS ¼fp=k½A ,d fljk 20°C ij nwljk fljk 60°C ij j[kk x;k gSA xeZ fljs }kjk izfr lsd.M yh x;h Å"ek dh ek=kk dh x.kuk dhft;sA ,Y;wfefu;e dh Å"ek pkydrk = 200 okWV/eh–°C rFkk rkacs dh Å"ek pkydrk = 390 okWV@eh–°C gSA
Ans.
2.36 J
25.
Figure shows an aluminium rod joined to a copper rod. Each of the rods has length of 20 cm and area of cross-section 0.20 cm2. The junction is maintained at a constant temperature 40°C and the two ends are maintained at 80°C. Calculate the amount of heat taken out from the cold junction in one minute after the steady state is reached. The conductivities are KAt = 200 W/m–°C and KCu = 400 W/m–°C. fp=k esa n'kkZ;k x;k gS fd ,d ,Y;wfefu;e dh NM+] rkacs dh NM+ ls tksM+ nh x;h gSA izR;sd NM+ dh yEckbZ 20 lseh rFkk vuqiLz Fk dkV dk {ks=kQy 0.20 lseh2 gSA laf/k dk rki 40°C rFkk nksuksa fljksa dk rki 80°C cuk;s j[kk tkrk gSA LFkk;h voLFkk izkIr djus ds i'pkr~ B.Mh laf/k ij 1 fefuV esa ckgj yh x;h Å"ek dh ek=kk dh x.kuk dhft;sA pkydrkvksa ds eku KAt = 200 okWV/eh–°C rFkk KCu = 400 okWV@eh–°C gSA
Ans.
144 J
26.
Consider the situation shown in figure. The frams is made of the same material and has a uniform crosssectional area everywhere. Calculate the amount of heat flowing per second through a cross-section of the bent part if the total heat taken out per second from the end at 100°C is 130 J.
fp=k esa iznf'kZr fLFkfr ij fopkj dhft;sA Ýse dk fuekZ.k leku inkFkZ ls gh fd;k x;k gS rFkk bldk vuqiLz Fk dkV izR;sd LFkku ij ,d leku gSa ;fn 100°C rki okys fljs ls izfr lsd.M yh x;h Å"ek 130 twy gS rks eqMs+ gq, Hkkx ds vuqiLz fk dkV ls izfr lsd.M izokfgr Å"ek dh ek=kk dh x.kuk dhft;sA
Ans.
60 J
27.
Suppose the bent part of the frams of the previous problem has a thermal conductivity of 780 J/m–s–°C whereas it is 390 J/m–s–°C for the straight part. Calculate the ratio of the rate of heat flow through the bent part to the rate of heat flow through the straight part. eku yhft;s fd fiNys iz'u esa Ýse dh Å"ek pkydrk 780 twy@eh–ls–°C, tcfd lh/ks Hkkx dh 390 twy@eh–ls–°C gSA
eqM+s gq, Hkkx ls Å"ek izokg dh nj rFkk lh/ks Hkkx ls Å"ek izokg dh nj ds vuqikr dh x.kuk dhft;sA Ans.
12 : 7
28.
A room has a window fitted with a single 1.0 m × 2.0 m glass of thickness 2 mm. (a) Calculate the rate of heat flow through the closed window when the temperature inside the room is 32°C and that outside is 40°C. (b) The glass is now replaced by two glasspanes, each having a thickness of 1mm and separated by a distance of 1 mm. Calculate the rate of heat flow under the same conditions of temperature. Thermal conductivity of window glass = 1.0 J/m–s–°C and that of air = 0.025 J/m–s–°C. manishkumarphysics.in
Page # 20
Chapter # 28
Heat Transfer
,d dejs dh f[kM+dh esa 2 feeh eksVk rFkk 1.0 eh × 2.0 eh okyk ,d vdsyk dk¡p yxk gqvk gSA (a) can f[kM+dh ls izokfgr Å"ek dh nj dh x.kuk dhft;s tcfd dejs ds vUnj rki 32°C rFkk ckgj dk rki 40°C gSA (b) dkap dks vc nks iryh dkap dh ijrks]a izR;sd dh eksVkbZ 1 feeh gS] ls izfrLFkkfir dj fn;k tkrk gS rFkk ijrksa ds chp dh nwjh 1 feeh gSA leku rkih; voLFkkvksa esa Å"ek izokg dh Å"ek pkydrk = 1.0 twy@eh–ls–°C rFkk ok;q dh Å"ek pkydrk = 0.025 twy@eh– ls–°C gSA Ans.
(a) 8000 J/s
29.
The two rods shown in figure have identical geometrial dimensions. They are in contact with two het baths at temperatures 100°C and 0°C. The temperature of the junction is 70°C. Find the temperature of the junction if the rods are interchanged. fp=k esa iznf'kZr nksuksa NM+kas dh T;kfefr; foek,¡ ,d leku gSA budks nks Å"ed uknksa ls tksM+ fn;k x;k gSA] ftuds rki 100°C ,oa 0°C gSA laf/k dk rki 70°C gSA ;fn NM+kas dks ijLij ifjofrZr dj fn;k tk;s rks laf/k dk rki Kkr dhft;sA
Ans.
30°C
30.
The three rods shown in figure have identical geometrical dimensions. Heat flows from the hot end at a rate of 40 W in the arrangement (a). Find the rates of heat flow when the rods are joined as in arrangement (b) and in (c) Thermal conductivities of aluminium and copper are 200 W/m–°C and 400 W/m–°C respectively. fp=k esa iznf'kZr dh x;h rhu NM+kas dh T;kferh; foek,¡ ,d leku gSA bl O;oLFkk esa xeZ fljs ls 40 okWV dh nj ls Å"ek izokfgr gks jgh gSA (a) ;fn NM+kas dks (b) vFkok (c) dh Hkkfr fn;k tk;s rks Å"ek izokg dh nj Kkr dhft;sA ,Y;wfefu;e rFkk rkacs dh Å"ek pkydrk,¡ Øe'k% 200 okWV@eh–°C rFkk 400 okWV/eh–°C gSA
Ans.
75 W, 400 W
31.
Four identical rods AB, CD, CF and DE are joined as shown in figure. The length, cross-sectional area and thermal conductivity of each rod are l, A and K respectively. The ends A, E and F are maintained at temperatures T1, T2 and T3 respectively. Assuming no loss of heat to the atmosphere, find the temperature at B. pkj ,d tSlh NM+s AB, CD, CF rFkk DE fp=k esa n'kkZ;s vuqlkj tksMh+ x;h gSA izR;sd NM+ dh yEckbZ vuqiLz FkdkV {ks=kQy ,oa Å"ek pkydrk Øe'k% , A rFkk K gSA fljs A, E rFkk F ds rki Øe'k% T1, T2 o T3 cuk;s j[ks tkrs gSAa ;g ekurs gq, fd okrkoj.k esa dksbZ Å"ek gkfu ugha gksrh gS] B dk rki Kkr dhft;sA
Ans. 32.
(b) 381 J/s
3T1 2(T2 T3 ) 7 Seven rods A, B, C, D, E, F and G are joined as shown in figure. All the rods have equal cross-sectional area A and length l. The thermal conductivities of the rods are KA = KC = K0, KB = KD = 2K0, KE = 3K0, KF = 4K0 and KG = 5K0. The rod E is kept at a constant temperature T1 and the rod G is kept at a constant temperature T2(T2 > T1). (a) Show that the rod F has a uniform temperature T = (T1 + 2T2)/3. (b) Find the rate of heat flow from the source which maintains the temperature T2. rki NM+s A, B, C, D, E, F ,oa G fp=kkuqlkj tksMh+ x;h gSA izR;sd NM+ dk vuqiLz Fk dkV {ks=kQy A rFkk yEckbZ gSA NM+kas dh Å"ek pkydrk,¡ KA = KC = K0, KB = KD = 2K0, KE = 3K0, KF = 4K0 rFkk KG = 5K0 gSA NM+ E dk fu;r rki T1 ij manishkumarphysics.in
Page # 21
Chapter # 28
Heat Transfer
,oa NM+ G dks fu;r rki T2(T2 > T1) ij j[kk x;k gSA (a) O;Dr dhft;s fd NM+ F dk le:i rki T = (T1 + 2T2) / 3 gksxkA (b) lzksr ls Å"ek izokg dh nj Kkr dhft;sA ftls T2 rki ij j[kk x;k gSA
Ans.
4K 0 A( T2 T1 ) 3l
33.
Find the rate of heat flow through a cross-section of the rod shown in figure (2 > 1). Thermal conductivity of the material of the rod is K. fp=k esa iznf'kZr NM+ dh vuqiLz Fk dkV ls izokfgr gksus okyh Å"ek dh nj Kkr dhft;s (2 > 1) NM+ ds inkFkZ dh Å"ek pkydrk K gSA
Ans.
(b)
34.
A rod of negligible heat capacity has length 20 cm, area of cross-section 1.0 cm2 and thermal conductivity 200 W/m–°C. The temperature of one end is maintained at 0°C and that of the other end is slowly and linearly varied from 0°C to 60°C in 10 minutes. Assuming no loss of heat through the sides, find the total heat transmitted through the rod in these 10 minutes. ux.; Å"ek /kkfjrk okyh ,d NM+ dh yEckbZ 20 lseh] vuqizLFk dkV 1.0 lseh2 rFkk Å"ek pkydrk 200 okWV@eh–°C gSA ,d fljs dk rki 0°C cuk;s j[kk tkrk gS rFkk nwljs fljs dk rki 0°C ls 60°C dj fn;k tkrk gSA ;g ekurs gq, fd fdukjksa ls dksbZ Å"ek gkfu ugha gksrh gS] bu 10 fefuV esa NM+ ls izokfgr dqy Å"ek Kkr dhft;sA 1800 J
Ans. 35.
Kr1r2 ( 2 1 ) L
A hollow metallic sphere of radius 20 cm surrounds a concentric metallic sphere of radius 5 cm. The space between the two spheres is filled with a nonmetallic material. The inner and outer spheres are maintained at 50°C and 10°C respectively and it is found that 100 J of heat passes from the inner sphere to the outer sphere per second. Find the thermal conductivity of the material between the spheres. 20 lseh f=kT;k okyk ,d [kks[kyk /kkfRod xksyk] 5 lseh f=kT;k okys ,d ldsUnzh; /kkfRod xksys dks ?ksjs gq, gSAa nksuksa xksyksa ds chp dk LFkku ,d v&/kkfRod inkFkZ ls Hkjk gqvk gSA vkarfjd ,oa cká xksyksa dk rki Øe'k% 50°C rFkk 10°C j[kk tkrk gS rFkk ;g izfs {kr fd;k tkrk gSA fd vkarfjd xksys ls cká xksys dh vksj izfr lsd.M 100 twy Å"ek izokfgr gksrh gSA xksyksa
ds chp esa Hkjs gq, inkFkZ dh Å"ek pkydrk Kkr dhft;sA Ans.
3.0 W/m–°C
36.
Figure shows two adiabatic vessels, each containing a mass m of water at different temperatures. The ends of a metal rod of length L, area of cross-section A and thermal conductivity K, are insertd in the water as shown in the figure. Find the time taken for the difference between the temperatures in the vessels to become half of the original value. The specific heat capacity of water is s. Nelgect the heat capacity of the rod and the container and any loss of heat to the atmosphere. fp=k esa nks :nks"e ik=k iznf'kZr fd;s x;s gS]a izR;sd esa ikuh dk m nzO;eku fHkUu&fHkUu rkiksa ij gSA L yEckbZ] A vuqiLz Fk dkV {ks=kQy rFkk K Å"ek pkydrk okyh ,d NM+ ds fljs fp=k esa n'kkZ;s vuqlkj ikuh esa Mky fn;s x;s gSAa ik=kksa esa rkikarj dk eku ewy rkikUrj dk vk/kk gksus esa yxk le; Kkr dhft;sA ikuh dh fof'k"V Å"ek s gSA NM+ rFkk ik=k dh Å"ek /kkfjrk ,oa
okrkoj.k dks gksus okyh Å"ek gkfu ux.; eku yhft;sA
Ans.
Lms In 2 2KA manishkumarphysics.in
Page # 22
Chapter # 28 Heat Transfer 37. Two bodies of masses m1 and m2 and specific heat capacities s1 and s2 are connected by a rod of length , cross-sectional area A, thermal conductivity K and negligible heat capacity. The whole system is thermally insulated. At time t = 0, the temperature of the first body is T1 and the temperature of the second body is T2(T2 > T1). Find the temperature difference between the two bodies at time t. m1 rFkk m2 nzO;ekuksa okyh rFkk s1 ,oa s2 fof'k"V Å"ekvksa okyh nks oLrq,¡ yEckbZ] A vuqiLz Fk dkV {ks=kQy ,oa K Å"ek pkydrk o ux.; Å"ek /kkfjrk okyh NM+ ls tksM+ nh x;h gSA lEiw.kZ fudk; Å"eh; :i ls vo:) gSA le; t = 0 ij izFke oLrq dk rki T1 rFkk nwljh oLrq dk rki T2(T2 > T1) gSA le; t ij nksuksa oLrqvksa dk rkikarj Kkr dhft;sA Ans. 38.
Ans. 39.
Ans. 40.
Ans. 41.
(T2 – T1) e–t where
KA(m1s1 m 2 s 2 ) lm1m 2 s1s 2
An amount n (in moles) of a monatomic gas at an initial temperature T0 is enclosed in a cylindrical vessel fitted with a light piston. The surrounding air has a temperature Ts(> T0) and the atmospheric pressure is pa. Heat may be conducted between the surrounding and the gas through the bottom of the cylinder. The bottom has a surface area A, thickness x and thermal conductivity K. Assuming all changes to be slow, find the distance moved by the piston in time t. ,d csyukdkj ik=k esa ,d gYdk fiLVu yxk gqvk gSA blesa n ek=kk esa (eksy es)a ,d ijek.kqd xSl Hkjh gqbZ gS] ftldk izkjfEHkd rki T0 gSA vklikl dh ok;q dk rki Ts(> T0) gS rFkk ok;qe.My dk nkc pa gSA flys.Mj ds iSnas ls gksdj okrkoj.k ,oa xSl ds chp Å"ek pkyu gks ldrk gSA iSnas dh i`"Bh; {ks=kQy A ] eksVkbZ x rFkk Å"ek pkydrk K gSA ;g eku yhft;s fd leLr ifjorZu cgqr /khjs gks jgs gS] le; t esa fiLVu }kjk pyh x;h nwjh Kkr dhft;sA
nR –2KAt/5 Rnx ) Pa A (Ts – T0) (1 – e Assume that the total surface area of a human body is 1.6 m2 and that it radiates like an ideal radiator. Calculate the amount of energy radiated per second by the body if the body temperature is 37°C. Stefan constant is 6.0 × 10–8 W/m2–K4. ;g eku yhft;s fd ekuo 'kjhj dk lEiw.kZ {ks=kQy 1.6 eh2 rFkk ;g ,d vkn'kZ fofdjd dh Hkkafr fofdfjr djrk gSA ;fn 'kjhj dk rki 37°C gS rks izfr lsd.M fofdfjr Å"ek dh ek=kk dh x.kuk dhft;sA LVhQu fu;rkad dk eku 6.0 × 10– 8 okWV@eh2–K4 887 J
Sol
Calculate the amount of heat radiated per second by a body of surface area 12 cm2 kept in thermal equilibrium in a room at temperature 20°C. The emissivity of the surface = 0.80 and = 6.0 × 10–8 W/m2–K4. 20°C rki okys dejs esa Å"eh; lkE;koLFkk dh fLFkfr esa fLFkr 12 lseh2 i`"Bh; {ks=kQy okyh oLrq ls izfr lsd.M fofdfjr Å"ek dh ek=kk dh x.kuk dhft;sA lrg dh mRltZdrk = 0.80 rFkk = 6.0 × 10–8 okWV/eh2–K4 0.42 J A solid aluminium sphere and a solid copper sphere of twice the radius are heated to the same temperature and are allowed to cool under identical surrounding temperatures. Assume that the emissivity of both the sphere is the same. Find the ratio of (a) the rate of heat loss from the aluminium sphere to the rate of heat loss from the copper sphere and (b) the rate of fall of temperature of the aluminium sphere to the rate of fall of temperature of the copper sphere. The specific heat capacity of aluminium = 900 J/kg–°C and that of copper = 390 J/kg–°C. The density of copper = 3.4 times the density of aluminium.
,d Bksl ,Y;wfefu;e dk xksyk vkSj nqxuq h f=kT;k okyk dkWij ds Bksl xksys dks ,d leku rkieku rd xeZ fd;k tkrk gS vkSj mUgsa ,d leku cká rkieku okys okrkoj.k esa B.Mk gksus fn;k tkrk gSA ;g ekfu, fd nksuksa xksyks dh mRltZdrk leku gSA rks Kkr djks (a) ,Y;wfefu;e ds xksys ls Å"ek gkfu dh nj dk dkWij xksys ds Å"ek gkfu dh nj ls vuqikr D;k gksxkA (b) ,Y;wfefu;e ds xksys dk rkieku fxjus dh nj dk dkWij xksys ds rkieku fxjus dh nj ls vuqikr D;k gksxkA ,Y;wfefu;e dh fof'k"V Å"ek /kkfjrk = 900 J/kg–°C vkSj dkWij dh fof'k"V Å"ek /kkfjrk = 390 J/kg–°C rFkk dkWij dk ?kuRo ] ,Y;wfefu;e ds ?kuRo dk 3.4 xquk gSA HCV_Ch-28_Ex_41 Solution on right side
Ans.
(a) 1 : 4
c rc sc (b) r s a a a 42.
Do 16.
A 100 W bulb has tungsten filament of total length 1.0 m and radius 4 × 10–5 m. The emissivity of the filament is 0.8 and = 6.0 × 10–8 W/m2–K4. Calculate the temperature of the filament when the bulb is operating at correct wattage. 100 okWV ds cYc esa VaxLVu ds rUrq dh dqy yEckbZ 1.0 eh- rFkk f=kT;k 4 × 10–5 eh gSA rarq dh mRltZdrk 0.8 rFkk = 6.0 × 10–8 okWV/eh2–K4 gSA tc cYc lgh 'kfDr ij dk;Z dj jgk gks rks rUrq ds rki dh x.kuk dhft;sA manishkumarphysics.in
Page # 23
gy.
Chapter # 28 Ans. 1700 K 43.
Ans. 44.
Ans. 45.
Ans. 46.
Ans. 47.
Ans. 48.
Heat Transfer
A spherical ball of surface area 20 cm2 absorbs any radiation that falls on it. It is suspended in a closed box maintained at 57°C. (a) Find the amount of radiation falling on the ball per second. (b) Find the net rate of heat flow to or from the ball at an instant when its temperature is 200°C. Stefan constant = 6.0 × 10–8 W/m2–K4. 20 lseh2 i`"B {ks=kQy okyh ,d xksykdkj xsna bl ij vkifrr leLr fofdj.k dk vo'kks"k.k dj ysrh gSA bldks ,d can izdks"B esa yVdk;k x;k gSA ftldk rki 57°C cuk;s j[kk tkrk gSA (a) xsna ij izfr lsd.M vkifrr fofdj.k dh ek=kk Kkr dhft;sA (b) ftl {k.k ij xsna dk rki 200°C gS rks bldh vksj vFkok blls gksus okys Å"ek izokg dh dqy nj Kkr dhft;sA LVhQu fu;rkad = 6.0 × 10–8 okWV/eh2–K4 (a) 1.4 J (b) 4.58 W from the ball A spherical tungsten piece of radius 1.0 cm is suspended in an evacuated chamber maintained at 300 k. The piece is maintained at 1000 K by heating it electrically. Find the rate at which the electrical energy must be supplied. The emissivity of tungsten is 0.30 and the stefan constant is 6.0 × 10–8 W/m2–K4. 300 k fLFkj rki okys fuokZfrr izdks"B esa VaxLVu dk 1.0 lseh f=kT;k okyk xksykdkj VqdM+k yVdk;k x;k gSA fo|qr /kkjk }kjk xeZ djds VqdM+s dk rki 1000 K fLFkj j[kk tkrk gSA Kkr dhft;s fdl nj ls fo|qr ÅtkZ iznku djuh gksxhA VaxLVu dh mRltZdrk 0.30 gS rFkk LVhQu dk fu;rkad dk eku = 6.0 × 10–8 okWV/eh2–K4 22 W A cubical block of mass 1.0 kg and edge 5.0 cm is heated to 227°C. It is kept in an evacuated chamber maintained at 27°C. Assuming that the block emits radiation like a blackbody, find the rate at which the temperature of the block will decrease. Specific heat capacity of the material of the block is 400 J/kg–K. 1.0 fdxzk nzO;eku rFkk 5.0 lseh Hkqtk okyk ,d ?kukdkj CykWd 227°C rd xeZ fd;k x;k gSA bldk ,d fuokZfrr izdks"B esa j[kk x;k gSA ftldk rki 27°C gSA ;g ekurs gq, fd CykWd] Ñf".kdk oLrq dh Hkkafr fofdj.k mRlftZr djrk gS] CykWd ds rki esa deh dh nj Kkr dhft;sA CykWd ds inkFkZ dh fof'k"V Å"ek 400 twy/fdxzk–K gSA 0.12°C/s A copper sphere is suspended in an evacuated chamber maintained at 300 K. The sphere is maintained at a constant temperature of 500 K by heating it electrically. A total of 210 W of electric power is needed to do it. When the surface of the copper sphere is completely blackened, 700 W is needed to maintain the same temperature of the sphere. Calculate the emisivity of copper. rkacs dk ,d xksyk 300 K rki okys fuokZfrr izdks"B esa yVdk;k x;k gSA fo|qr fof/k ls xeZ djds xksys dk rki 500 K fLFkj j[kk tkrk gSA blds fy;s dqy 210 okWV fo|qr 'kfDr dh vko';drk gksrh gSA tc rkacs ds xksys dh lEiw.kZ lrg dkyh dj nh tkrh gS] xksys dk ;gh rki cuk;s j[kus ds fy;s 700 okWV dh vko';drk gksrh gSA rkacs dh mRltZdrk dh x.kuk dhft;sA 0.3 A spherical ball A of surface area 20 cm2 is kept at the centre of a hollow spherical shell B of area 80 cm2. The surface of A and the inner surface of B emit as blackbodies. Assume that the thermal conductivity of the material of B is very poor and that of A is very high and that the air between A and B has pumped out. The heat capacities of A and B are 42 J/°C and 82 J/°C respectively. Initially, the temperature of A is 100°C and that of B is 20°C. Find the rate of change of temperature of A and that of B at this instant. Explain the effects of the assumptions listed in the problem. 20 lseh2 i`"Bh; {ks=kQy okyh ,d xksykdkj xsna A, 80 lseh2 i`"Bh; {ks=kQy okyh [kks[kyh xksykdkj xsna B ds dsUnz ij j[kh gqbZ gSA A dh lrg rFkk B dh vkarfjd lrg Ñf".kdk oLrq dh Hkkafr mRltZu djrh gSA eku yhft;s fd B ds inkFkZ dh Å"ek pkydrk cgqr gh de gS rFkk A ds inkFkZ dh cgqr vf/kd gS ,oa A o B ds chp dh gok iEi ls ckgj fudky nh x;h gSA A o B dh Å"ek /kkfjrk,¡ Øe'k% 42 twy/°C vkSj 82 J/°C gSA izkjEHk es]a A dk rki 100°C rFkk B dk 20°C gSA bl {k.k ij A rFkk B ds rki ifjorZu dh nj Kkr dhft;sA bl leL;k esa nh x;h /kkj.kkvksa ds izHkkoksa dh O;k[;k dhft;sA 0.05°C/s and 0.01°C/s A cylindrical rod of length 50 cm and cross-sectional area 1 cm2 is fitted between a large ice chamber at 0°C and an evacuated chamber maintained at 27°C as shown in figure. Only small portions of the rod are inside the chambers and the rest is thermally insulated from the surrounding. The cross-section going into the evacuated chamber is blackened so that it completely absorbs any radiation falling on it. The temperature of the blackened end is 17°C when steady state is reached. Stefan constant = 6 × 10–8W/m2–K4. Find the thermal conductivity of the material of the rod. tSlk fd fp=k esa iznf'kZr gS] 0°C rki okys cQZ ds d{k rFkk 27°C rki okys ,d fuokZfrr izdks"B ds chp esa ,d csyukdkj NM+ tksMh+ x;h gS] NM+ dh yEckbZ 50 lseh rFkk blds vuqiLz Fk dkV dk {ks=kQy 1 lseh2 gSA NM+ ds vR;Yi Hkkx gh d{kksa ds
vUnj gS rFkk 'ks"k Hkkx okrkoj.k ls Å"eh; :i ls foyfxr gSA fuokZfrr d{k esa izfo"V vuqiLz Fk dkV dkyk fd;k x;k gSA ftlls ;g bl ij vkifrr gksus okys fofdj.k dks iw.kZr;k vo'kksf"kr dj ysrk gSA lkE;koLFkk izkIr djus ij dkys fd;s x;s manishkumarphysics.in
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Chapter # 28
Heat Transfer
fljs dk rki 17°C gks tkrk gSA LVhQu fu;rkad = 6 × 10–8 okWV/eh2–K4 A NM+ ds inkFkZ dh Å"ek pkydrk Kkr dhft;sA
Ans.
1.8 W/m–°C
49.
One end of a rod of length 20 cm is inserted in a furnace at 800 K. The sides of the rod are covered with an insulating material and the other end emits radiation like a blackbody. The temperature of this end is 750 K in the steady state. The temperature of the surrounding air is 300 K. Assuming radiation to be the only important mode of energy transfer between the surrounding and the open end of the rod, find the thermal conductivity of the rod. Stefan constant = 6.0 × 10–8 W/m2–K4. 20 lseh yEch NM+ dk ,d fljk 800 K rki okyh ,d HkV~Vh esa Mkyk x;k gSA NM+ ds fdukjs ,d Å"ekjks/kh inkFkZ ls
Ans. 50.
Ans. 51.
A calorimeter of negligible heat capacity contains 100 cc of water at 40°C. The water cools to 35°C in 5 minutes. The water is now replaced by K-oil of equal volume at 40°C. Find the time taken for the temperature to become 35°C under similar conditions. Specific heat capacities of water and K-oil are 4200 J/kg–K and 2100 J/kg–K respectively. Density of K-oil = 800 kg/m3. ux.; Å"ek /kkfjrk okys ,d dSyksjhehVj esa 40°C rki okyk 100 ?ku lseh ty gSA 5 fefuV esa 35°C rd B.Mk gks tkrk gSA ty ds LFkku ij vc 40°C rki ij dSjkslhu rsy ds leku vk;ru fy;k tkrk gSA leku ifjfLFkfr;ksa esa 35°C rki gksus esa fy;k x;k le; Kkr dhft;sA ty rFkk dSjkslhu rsy dh fof'k"V Å"ek,¡ Øe'k% 4200 twy/fdxzk–K gS rFkk 2100 twy/ fdxzk–K gSA dSjkslhu rsy dk ?kuRo 800 fdxzk-eh3 A 2 min A body cools down from 50°C to 45°C in 5 minutes and to 40°C in another 8 minutes. Find the temperature of the surrounding. ,d oLrq 5 fefuV esa 50°C ls 45°C rd B.Mh gksrh gS rFkk vxys 8 fefuV esa 40°C rd B.Mh gksrh gSA okrkoj.k dk rki
Kkr dhft;sA Ans.
34°C
52.
A calorimeter contains 50 g of water at 50°C. The temperature falls to 45°C in 10 minutes. When the calorimeter contains 100 g of water at 50°C, it takes 18 minutes for the temperature to become 45°C. Find the water equivalent of the calorimeter.
sheet Question
,d dSyksjhehVj esa 50 xzke ikuh] 50°C ij Hkjk gqvk gSA 10 feuV esa rkieku fxjdj 45°C gks tkrk gSA tc dSyksjhehVj esa 100 xzke ikuh 50°C ij Hkjk gS rks bldk rkieku 45°C gksus esa 18 feuV dk le; yxrk gSA dSyksjhehVj dk ty&rqY;kad crkb, \ HCV_Ch-28_Ex._52 Ans.
12.5 g
53.
A metal ball of mass 1 kg is heated by means of a 20 W heater in a room at 20°C. The temperature of the ball becomes steady at 50°C. (a) Find the rate of loss of heat to the surrounding when the ball is at 50°C. (b) Assuming Newton’s law of cooling, calculate the rate of loss of heat to the surrounding when the ball is at 30°C. (c) Assume that the temperature of the ball rises uniformly from 20°C to 30°C in 5 minutes. Find the total loss of heat to the surrounding during this period. (d) Calculate the specific heat capacity of the metal. fdlh dejs esa 1 fdxzk nzO;eku okyh /kkrq dh ,d xsna 20°C rki ij 20 okWV ds fo|qr ghVj }kjk xeZ dh tkrh gSA xsna dk rki 50°C ij fLFkj gks tkrk gSA (a) tc xsna 50°C ij gS rks okrkoj.k esa Å"ek gkfu dh nj Kkr dhft;sA (b) eku yhft;s fd U;wVu dk 'khryu dk fu;e ykxw jgrk gS] tc xsna 30°C rki ij gS] tc xsna 30°C rki ij gS] okrkoj.k esa gksus okyh Å"ek gkfu dh nj dh x.kuk dhft;sA (c) eku yhft;s fd 5 fefuV esa xsna dk rki ,d leku :i ls c<+dj 20°C ls 30°C gks tkrk gS] bl dky esa okrkoj.k esa gksus okyh Å"ek gkfu Kkr dhft;sA (d) /kkrq dh fof'k"V Å"ek Kkr dhft;sA
Ans.
(a) 20 W
54.
20 W (c) 1000 J (d) 500 J/kg–K 3 A metal block of heat capacity 80 J/°C placed in a room at 20°C is heated electrically. The heater is switched off when the temperature reaches 30°C. The temperature of the block rises at the rate of 2°C/s just after the heater is switched on and falls at the rate of 0.2 °C/s just after the heater is switched off. Assume Newton’s law of cooling to hold. (a) Find the power of the heater. (b) Find the power radiated by the block just after the heater is switched off. (c) Find the power radiated by the block when the temperature of the block is 25°C. (d)
(b)
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Chapter # 28 Heat Transfer Assuming that the power radiated at 25°C respresents the average value in the heating process, find the time for which the heater was kept on. HCV_Ch-28_Sub._54 80 J/°C Å"ek /kkfjrk okys ,d /kkrq xqVds dks 20°C okys ,d dejs esa j[kdj fo|qr :i ls xeZ fd;k tkrk gSA tc rkieku 30°C igqap tkrk gS rks ghVj dks can dj fn;k tkrk gSA tc ghVj dks 'kq: djus ds Bhd ckn xqVds dk rkieku 2°C/s dh nj ls c<+rk gS vkSj tc ghVj dks can djus ds Bhd ckn xqVds dk rkieku 0.2 °C/s dh nj ls ?kVrk gSA ;g ekfu, fd U;wVu dk 'khrry dk fu;e ;gka ij ykxw gSA (a) ghVj dh 'kfDr crkb,A (b) ghVj dks can djus ds Bhd ckn xqVds }kjk fodfjr (radiated) 'kfDr crkb,A (c) tc xqVds dk rkieku 25°C gS rc xqVds ds }kjk fodfjr (radiated) 'kfDr crkb,A (d) ;g ekurs gq, fd 25°C ij fofdfjr 'kfDr Å"eh; izfØ;k esa vkSlr eku dks crkrk gS rks og le; crkb, ftrus le;
rd ghVj dks pkyw j[kk x;kA
Ans.
(a) 160 W (c) 8 W
(b) 16 W (d) 5.2 s
55.
A hot body placed in a surrounding of temperature 0 obeys Newton’s law of cooling
d = – k( – 0). Its dt temperature at t = 0 is 1. The specific heat capacity of the body is s and its mass is m. Find (a) the maximum heat that the body can lose and (b) the time starting from t = 0 in which it will lose 90% of this maximum heat.
0 rki
okys okrkoj.k esa j[kh gqbZ ,d xeZ oLrq U;wVu ds 'khryu ds fu;e
d = – k( – 0) dk dt
ikyu djrh gSA t = 0
ij bldk rki 1 gSA oLrq dh fof'k"V Å"ek s rFkk nzO;eku m gSA Kkr dhft;s % (a) og vf/kdre Å"ek tks oLrq [kks ldrh gS rFkk (b) t = 0 ls izkjEHk djds bl vf/kdre Å"ek dk 90% [kksus esa yxk le;A Ans.
(a) ms(1 – 0)
(b)
In 10 k
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