calorimetry

Chapter # 25

Calorimetry

SOLVED EXAMPLES

EXAMPLE 25.1 What is the kinetic energy of a 10 kg mass moving at a speed of 36 km/h in calorie ? Sol.

1 1 mv 2 = × 10 kg × 2 2 = 500 J =

 36  10 3 m     3600 s   

2

500 cal  120 cal. 4.186

EXAMPLE 25.2 A copper block of mass 60 g is heated till its temperature is increased by 20 0 C. Find the heat capacity of copper = 0.09 cal/g– 0C . Sol. The heat supplied is Q = ms = (60g) (0.09 cal/g–0C) (200 C) = 180 cal. The quantity ms is called the heat capacity of the body. Its unit is J/K. The mass of water having the same heat capacity as a given body is called the water equivalent of the body. EXAMPLE 25.3 A piece of ice of mass 100 g and at a temperature 00C is put in 200 g of water at 250 C. How much ice will melt as the temperature of the water reaches 00 C ? The specific heat capacity of water = 4200 J/kg– K and the specific latent heat of fusion of ice = 3.4 ×105 J/kg. Sol. The heat released as the water cools drown from 250 C to 00 C is Q = ms = (0.2 kg) (4200 J/kg–K)(25 K)= 21000 J. The amount of ice melted by this much heat is given by m=

21000 J Q = = 62 g. 3.4  10 5 J/kg L

EXAMPLE 25.4 A calorimeter of water equivalent 15 g contains 165 g of water at 250C . Steam at 1000 C is passed through the water for some time . The temperature is increased to 300C and the mass of the calorimeter and its contents is increased by 1.5 g. Calculate the specific latent heat of vaporization of water.Specific heat capacity of water is 1 cal/g– 0C. Sol. Let L be the specific latent heat of vaporization of water. The mass of the steam condensed is 1.5 g. Heat lost in condensation of steam is Q1 = ( 1.5 g ) L. The condensed water cools from 1000 C to 300 C. Heat lost in this process is Q2 = ( 1.5 g ) ( 1 cal/g– 0C) ( 700C) = 105 cal. Heat supplied to the calorimeter and to the cold water during the rise in temperature from 250 C to 300 C is Q3 = ( 15 g + 165 g ) ( 1 cal/g– 0 C ) ( 50 C) = 900 cal. If no heat is lost to the surrounding, ( 1.5 g ) L + 105 cal = 900 cal or, L = 530 cal/g.

QUESTIONS

FOR

SHORT

ANSWER

Q.1

Is heat a conserved quantity?

Q.2

The calorie is defined as 1 cal = 4.186 joule. Why not as 1 cal = 4J to make the conversion easy?

Q.3

A calorimeter is kept in a wooden box to insulate it thermally from the surroundings. Why is it necessary?

Q.4

In a calorimeter, the heat given by the hot object is assumed to be equal to the heat taken by the cold object. manishkumarphysics.in

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Chapter # 25 Calorimetry Does it means that heat of the two objects taken together remains constant? Q.5

In regnault’s apparatus for measuring specific heat capacity of a solid, there is an inlet and an outlet in the steam chamber. The inlet is near the top and the outlet is near the bottom. Why is it better than the opposite choice where the inlet is near the bottom and the outlet is near the top?

Q.6

When a solid melts or a liquid boils, the temperature does not increase even when heat is supplied. Where does the energy go? [HCV_Chp.25_Short Ans._6]

Q.7

What is the specific heat capacity of (a) melting ice (b) boiling water? [HCV_Chp.25_Short Ans._7]

Q.8

A person’s skin is more severely burnt when put in contact with 1 g of steam at 100ºC than when put in contact with 1 g of water at 100ºC. Explain.

Q.9

The atmospheric temperature in the cities on sea-coast change very little. explain ?

Q.10

Should a thermometer bulb have large heat capacity or small heat capacity ? [HCV_Chp.25_Short Ans._10]

Objective - I 1.

The specific heat capacity of a body depends on (A) the heat given (B) the temperature raised (C) the mass of the body (D*) the material of the body fdlh oLrq dh fof'k"V Å"ek fuHkZj djrh gS (A) nh x;h Å"ek ij (B) rki ij o`f) ij (C) oLrq ds nzO;eku ij (D*) oLrq ds inkFkZ ij

2.

Water equivalent of a body in measured in (A*) kg (B) calorie

(C) kelvin

(D) m3

fdlh oLrq dk ty rqY;kad fuEu esa ekik tkrk gS (A*) fdxzk (B) dSyksjh (C) dsfYou

(D) eh3

3.

When a hot liquid is mixed with a cold liquid, the temperature of the mixture (A) first decreases then becomes constant (B) first increases then becomes constant (C) continuously increases (D*) is undefined for some time and then becomes nearly constant tc ,d xeZ nzo fdlh 'khry nzo esa feyk;k tkrk gS] rks feJ.k dk rki (A) igys de gksrk gS fQj fu;r gks tkrk gSA (B) igys c<+rk gS fQj fu;r gks tkrk gSA (C) fujUrj c<+rk gSA (D*) dqN le; ds fy;s vifjHkkf"kr jgrk gS rRi'pkr~ yxHkx fu;r jgrk gSA

4.

Which of the following pairs represent units of the same physical quantity ? (A) kelvin and joule (B) kelvin and calorie (C) newton and calorie (D*) joule and calorie fuEu esa ls dkSuls ;qXe ,d gh HkkSfrd jkf'k dh bdkbZ;ksa dks O;Dr djrs gSa (A) dsfYou rFkk twy (B) dsfYou rFkk dSyksjh (C) U;wVu rFkk dSyksjh (D*) twy rFkk dSyksjh

5.

Which of the following pairs of physical quantities may be represented in the same unit ? (A) heat and temperature (B) temperature and mole (C*) heat and work (D) specific heat and heat HkkSfrd jkf'k;ksa ds ;qXeksa esa ls dkSulk ,d gh bdkbZ }kjk O;Dr fd;k tk ldrk gS (A) Å"ek rFkk rki (B) rki rFkk eksy (C*) Å"ek rFkk dk;Z (D) fof'k"V Å"ek rFkk Å"ek

6.

Two bodies at different temperatures are mixed in a calorimeter. Which of the following quantities remains conserved ? (A) sum of the temperature of the two bodies (B) total heat of the two bodies (C*) total internal energy of the two bodies (D) internal energy of each body fHkUu&fHkUu rkiksa okyh nks oLrq,¡ ,d dSyksjh ekih ls feyk;h tkrh gSA fuEu jkf'k;ksa esa ls dkSulh lajf{kr jgsxh (A) nksuska oLrqvksa ds rkiksa dk ;ksx (B) nksuksa oLrqvksa dh dqy Å"ek (C*) nksuksa oLrqvksa dh dqy vkarfjd ÅtkZ (D) izR;sd oLrq dh vkarfjd ÅtkZ manishkumarphysics.in

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Chapter # 25 7.

Calorimetry

The mechanical equivalent of heat (A) has the same dimension of heat (C) has the same dimension as energy Å"ek dk ;kaf=kd&rqY;kad (A) dh foek,¡ Bhd Å"ek tSlh gksrh gSA (C) dh foek,¡ Bhd ÅtkZ tSlh gksrh gSA

(B) has the same dimension as work (D*) is dimensionless (B) dh foek,¡ Bhd dk;Z (D*) foekghu gksrk gSA

tSlh gksrh gSA

Objective - II 1.

The heat capacity of a body depends on (A) the heat given (C*) the mass of the body fdlh oLrq dh Å"ek/kkfjrk fuHkZj djrh gS (A) nh x;h Å"ek ij (C*) oLrq ds nzO;eku ij

(B) the temperature raised (D*) the material of the body (B) rki&o`f) ij (D*) oLrq ds inkFkZ

ij

2.

The ratio of specific heat capacity to molar heat capacity of a body (A) is a universal constant (B) depends on the mass of the body (C*) depends on the molecular weight of the body (D) is dimensionless fdlh oLrq dh fof'k"V Å"ek rFkk eksyj fof'k"V Å"ek dk vuqikr (A) ,d lkoZf=kd fu;rkad gksrk gS (B) oLrq ds nzO;eku ij fuHkZj djrk gSA (C*) oLrq ds xzke v.kqHkkj ij fuHkZj djrk gSA (D) foekghu gksrk gSA

3.

If heat is supplied to a solid, its temperature (A) must increase (B*) may increase

;fn fdlh Bksl dk Å"ek iznku dh tk;s] rks bldk rki (A) fuf'pr :i ls c<+sxk (B*) c<+ ldrk gSA 4.

5.

6.

(C*) may remain constant (C*) fu;r jg ldrk gSA

(D) may decrease (D) de

gks ldrk gSA

The temperature of a solid object is observed to be constant during a period. In this period (A*) heat may have been supplied to the body (B*) heat may have been extracted from the body (C) no heat is supplied to the body (D) no heat is extracted from the body fdlh fuf'pr dky esa ,d Bksl oLrq dk rki fu;r jgrk gSA bl dky esa (A*) ;g gks ldrk gS fd oLrq dks Å"ek iznku dh tk jgh gksA (B*) ;g gks ldrk gS fd oLrq ls Å"ek yh tk jgh (C) oLrq dks dksbZ Å"ek ugha nh tk jgh gksA (D) oLrq ls dksbZ Å"ek ugha yh tk jgh gksA The temperatuer of an object is observed to rise in a period. In this period (A) heat is certainly supplied to it (B) heat is certainly not supplied to it. (C*) heat may have been supplied to it (D*) work may have been done on it. fdlh dky esa ,d oLrq ds rki esa o`f) izfs {kr dh tkrh gSA bl dky esa (A) fuf'pr :i ls bldks Å"ek nh tk jgh gSA (B) fuf'pr :i ls bldks Å"ek nh tk jgh (C*) gks ldrk gS fd bldks Å"ek nh tk jgh gks (D*) gks ldrk gS fd bl ij dk;Z fd;k tk Heat and work are equivalent. This means, (A) when we supply heat to a body we do work on it (B) when we do work on a body we supply heat to it (C*) the temperature of a body can be increased by doing work on it. (D) a body kept at reat may be set into motion along a line by supplying heat to it. Å"ek rFkk dk;Z rqY; gSA bldk rkRi;Z gS (A) tc ge fdlh oLrq dks Å"ek iznku djrs gSa] ge bl ij dk;Z djrs gSaA (B) tc ge fdlh oLrq ij dk;Z djrs gSa] ge oLrq dks Å"ek iznku djrs gSaA (C*) oLrq ij dk;Z djds bldk rki c<+k;k tk ldrk gSA (D) fojkekoLFkk esa fLFkr oLrq dks Å"ek iznku djds] bldk ljy js[kk ds vuqfn'k xfr'khy fd;k

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gksA

gksA jgk gksA

tk ldrk gSA

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Chapter # 25

1.

Calorimetry

WORKED OUT EXAMPLES

Calculate the amount of heat required to convert 1.00kg of ice at - 10ºC into steam at 100ºC at normal pressure. Specific heat capacity of ice = 2100 J/kg-K, latent heat of fusion of ice = 3.36 × 105 J/kg, specific heat capacity of water = 4200 J/kg-K and latent heat of vaporization of water = 2.25 × 106 J/kg.

Sol.

Heat required to take the ice from - 10ºC to 0ºC = (1 kg) (2100 J/kg-K) (10 K) = 21000 J. Heat required to melt the ice at 0ºC to water = (1 kg) (3.36 × 105 J/kg) = 336000 J. Heat required to take 1 kg of water from 0ºC to 100ºC = (1 kg) (4200 J/kg)-K (100 K) = 420000 J. Heat required to convert 1 kg of water at 100ºC into steam = (1kg) (2.25 × 106 J/kg) = 2.25 × 106 J. Total heat required = 3.03 × 106 J.

2.

A 5 g piece of ice at - 20ºC is put into 10 g of water at 30ºC. Assuming that heat is exchanged only between the ice and the water, find the final temperature of the mixture. Specific heat capacity of ice = 2100 J/kg-ºC, specific heat capacity of water = 4200 J/kg-ºC and latent heat of fusion of ice = 3.36 × 105 J/kg.

Sol.

The heat given by the water when it cools down form 30ºC to 0ºC is (0.01 kg) (4200 J/kg-ºC) (30ºC) = 1260 J. The heat required to bring the ice to 0ºC is (0.005kg) (2100 J/kg-ºC) (20ºC) = 210 J. The heat required to melt 5 g of ice is (0.005 kg) (3.36 × 105 J/kg) = 1680 J. We see that whole of the ice cannot be melted as the required amount of heat is not provided by the water. Also, the jeat enough to bring the ice to 0ºC. Thus the final temperature of the mixutre is 0ºC with some of the ice melted.

3.

An aluminium container of mass 100 g contains 200g of ice at - 20ºC. Heat is added to the system at a rate of 100 cal/s. What is the temperature of the system after 4 minutes? Draw a rough sketch showing the variation in the temperature of the system as a function of the time. Specific heat capacity of ice = 0.5 cal/gºC, specific heat capacity of aluminium = 0.2 cal/g-ºC, specific heat capacity of water = 1 cal/g-ºC and latent heat of fusion of ice = 80 cal/g.

Sol.

Total heat supplied to the system in 4 minutes is Q = 100 cal/s × 240s = 2.4 × 104 cal. The heat required to take the system from - 20ºC to 0ºC = (100g) × 0.2 cal/g-ºC) × (20ºC) + (200 g) × (0.5 cal/g-ºC) × (20ºC) = 400 cal + 2000 cal = 2400 cal. 2400 s = 24 s. 100 The heat required to melt the ice at 0ºC = (200 g) (80 cal/g) = 16000 cal.

The time taken in this process =

16000 s = 160 s. 100 If the final temperature is , the heat required to take the system to the final temperature is = (100 g) (0.2 cal/g-ºC)  + (200 g) (1 cal/g-ºC) . Thus, 2.4 × 104 cal = 2400 cal + 16000 cal + (220 cal/ºC) 

The time taken in this process =

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Chapter # 25 or

Calorimetry



5600 cal 220 cal /º C = 25.5ºC.

The variation of the temperature as a function of time is sketched in figure.

4.

Sol.

A thermally isolated vessel contains 100 g of water at 0ºC. When air above the water is pumped out, some of the water freezes and some evaporates at 0ºC itself. Calculate the mass of the ice formed if no water is left in the vessel. Latent heat of vaporization of water at 0ºC = 2.10 × 106 J/kg and latent heat of fusion of ice = 3.36 × 105 J/kg. Total mass of the water = M = 100 g. Latent heat of vaporization of water at 0ºC = L1 = 21.0 × 105 J/kg. Latent heat of fusion of ice = L2 = 3.36 × 105 J/kg. Suppose, the mass of the ice formed = m. Then the mass of water evaporated = M - m. Heat taken by the water to evaporate = (M - m) L1 and heat given by the water in freezing = mL2 Thus, mL2 = (M-m) L1 or,

m

ML1 L1  L 2 (100 g) (21.0  10 5 J / kg) (21.0  3.36 )  10 5 J / kg

5.

Sol.

= 86 g

A lead bullet penetrates into a solid object and melts. Assuming that 50% of its kinetic energy was used to heat it, calculate the initial speed of the bullet. The initial temperature of the bullet is 27ºC and its melting point is 327ºC. Latent heat of fusion of lead = 2.5 × 104 J/kg and specific heat capacity of lead = 125 J/kgK. Let the mass of the bullet = m. Heat required to take the bullet from 27ºC to 327ºC = m × (125 J/kg-K) (300 K) = m × (3.75 × 104 J/kg) Heat required to melt the bullet = m × (2.5 × 104 J/kg). If the initial speed be , the kinetic energy is Thus, or

6.

Sol.

1 1 1 2 1 2 mv2 and hence the heat developed is  mv  = mv . 22  4 2

1 mv 2 = m (3.75 + 2.5) × 104 J/kg 4 = 500 m/s.

A lead ball at 30º is dropped from a height of 6.2km. The ball is heated due to the air resistance and it completely melts just before reaching the ground. The molten substance falls slowly on the ground. Calculate the latent heat of fusion of lead. Specific heat capacity of lead = 126 J/kg–ºC and melting point of lead = 330ºC. Assume that any mechanical energy lost is used to heat the ball. Use g = 10 m/s2. The initial gravitational potential energy of the ball = mgh = m × (10 m/s2) × 6.2 × 103m) = m × (6.2 × 104 m2/s2) = m × (6.2 × 104 J/kg). manishkumarphysics.in

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Chapter # 25 Calorimetry All this energy is used to heat the ball as it reaches the ground with a small velocity. Energy required to take the ball from 30ºC to 330ºC is m × (126 J/kg–ºC) × (300ºC) = m × 37800 J/kg and energy required to melt the ball at 330ºC = mL where L = latent heat of fusion of lead. Thus, m × (6.2 × 104 J/kg) = m × 37800 J/kg + mL or, L = 2.4 × 104 J/kg.

EXERCISE 1.

An aluminium vessel of mass 0.5kg constants 0.2 kg of water at 20ºC. A block of iron of mass 0.2 kg at 100ºC is gently put into the water. Find the equilibrium temperature of the nuxture. Specific heat capacities of aluminium, iron and water are 910 J/kg–K, 470 J/kg–K and 4200 J/kg–K respectively. 0.5 fdxzk nzO;eku okys ,Y;wfefu;e ds ik=k esa 20ºC ij ikuh 0.2 fdxzk ikuh gSA 0.2 fdxzk nzO;eku ,oa 100ºC rki okyk

yksgs dk ,d xqVdk /khjs ls ikuh esa NksM+ fn;k tkrk gSA lkE;koLFkk esa feJ.k dk rki Kkr dhft;sA ,Y;wfefu;e yksgs rFkk ty dh fof'k"V Å"ek,¡ Øe'k% 910 twy@fdxzk–K, 470 twy@fdxzk–K ,oa 4200 twy@fdxzk–K gSA Ans.

25ºC

2.

A piece of iron of mass 100g is kept inside a furnace for a long time and then put in a calorimeter of water equivalent 10 g containing 240 g of water at 20ºC. The mixture attains an equilibrium temperature of 60ºC. Find the temperature of the furnace. Specific heat capacity of iron = 470 J/kg-ºC. 100 xzke nzO;eku dk yksgs dk VqdM+k yEcs le; rd HkV~Vh esa j[kus ds i'pkr~] 10 xzke ty rqY;kad okys dSyksjh ekih esa Hkjs gq, 20ºC rki okys 240 xzke ty esa Mkyk tkrk gSA lkE;koLFkk izkIr djus ij feJ.k dk rki 60ºC gks tkrk gSA HkV~Vh dk rki Kkr dhft;sA yksgs dh fof'k"V Å"ek = 470 twy@fdxzk-ºC 950ºC

Ans. 3.

Ans. Q.4

The temperatures of equal masses of three different liquids A, B and C are 12ºC, 19ºC and 28ºC respectively. The temperature when A and B are mixed is 16ºC, and when B and C are mixed, it is 23ºC. What will be the temperature when A and C are mixed ? rhu leku nzO;eku okys vyx&vyx nzoksa A, B rFkk C ds fy, dk rkieku Øe'k% 12ºC, 19ºC rFkk 28ºC gSA tc A rFkk B dks fefJr fd;k tkrk gS rks rkieku 16ºC gS rFkk tc B ,oe~ C dks fefJr fd;k tkrk gS rks rkieku 23ºC gSA A rFkk C dks fefJr djus ij rkieku D;k gksxk ? HCV_Ch-25_Ex_3 20.3ºC Four 2cm × 2cm × 2cm cubes of ice are taken out from a refrigerator and are put in 200 ml of a drink at 10ºC. (a) Find the temperature of the drink when thermal equilibrium is attained in it. (b) If the ice cubes do not melt completely, find the amount melted. Assume that no heat is lost to the outside of the drink and that the container has negligible heat capacity. Density of ice = 900 kg/m3, density of the drink = 1000 kg/m3, specific heat capacity of the drink = 4200 J/kg-K, latent heat of fusion of ice = 3.4 × 105 J/kg. 2 lseh × 2 lseh × 2 lseh okys pkj vkbl D;wc jsfÝtjsVj ls fudky dj 10ºC rki okys 200 feyh is; esa Mkys tkrs gSaA (a) Å"eh; lkE;koLFkk izkIr gksus ds i'pkr~ is; dk rki Kkr dhft;sA (b) ;fn vkbl D;wc iw.kZr;k fi?kyrs ugha gS] rks fi?kyh

gqbZ ek=kk Kkr dhft;sA eku yhft;s fd is; ls ckgj Å"ek gkfu ugha gksrh gS rFkk ik=k dh Å"ek /kkfjrk ux.; gSA cQZ dk /kuRo = 900 fdxzk/eh3, is; dk ?kuRo = 1000 fdxzk/eh3, is; dh fof'k"V Å"ek = 4200 twy@fdxzk-K, cQZ ds xyu dh xqIr Å"ek = 3.4 × 105 twy@fdxzk A Ans.

(a) 0ºC

(b) 25 g

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Chapter # 25 Calorimetry Q.5 Indian style of colling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates. Most of the energy needed for evaporation is taken from the water itself and the water is cooled down. Assume that a pitcher contains 10 kg of water and 0.2 g of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decreases by 5ºC. Specific heat capacity of water = 4200 J/kg–ºC and latent heat of vaporization of water = 2.27 × 106 J/kg.

ihus dk ikuh B.Mk djus dh Hkkjrh; fof/k es]a ikuh dks ljU/kz nhokjksa okys eVds esa Hkjdj j[kk tkrk gSA ikuh /khjs&/khjs ckgj okyh lrg ij vkdj okf"ir gksrk jgrk gSA ikuh ds ok"iu ds fy;s vko';d Å"ek dk Hkkx Lo;a ikuh ls gh fy;k tkrk gS rFkk ikuh B.Mk gksrk jgrk gSA eku yhft;s fd eVds esa 10 fdxzk ikuh gS rFkk izfr lsd.M 0.2 xzke ikuh ckgj vkrk gSA ;g ekurs gq, fd okrkoj.k ls dksbZ Å"ek ikuh dks ugha ykSVkbZ tkrh gS] ml le; dh x.kuk dhft;s ftlds i'pkr~ rki esa 5ºC deh gks tk;sxhA ty dh fof'k"V Å"ek = 4200 twy/fdxzk–ºC ,oa ty ds ok"iu dh xqIr Å"ek = 2.27 × 106 twy/fdxzk Ans.

7.7 min

Q.6

A cube of iron (density = 8000 kg/m3, specific heat capacity = 470 J/kg-K) is heated to a high temperature and is placed on a large block of ice at 0ºC. The cube melts the ice below it, displaces the water and sinks. In the final equilibrium position, its upper surface just goes inside the ice. Calculate the initial temperature of the cube. neglect any loss of heat outside the ice and the cube. The density of ice = 900 kg/m3 and the latent heat of fusion of ice = 3.36 × 105 J/kg. yksgs ¼?kuRo = 8000 fdxzk@eh3, fof'k"V Å"ek = 470 twy@fdxzk-K) ds ,d ?ku dk mPp rki rd xeZ djds 0ºC rki okys

cQZ ds ,d cgqr cM+s CykWd ij j[k fn;k tkrk gSA ?ku vius uhps okys cQZ dks fi?kyk nsrk gS] ikuh dks foLFkkfir djds cgk nsrk gSA vafre lkE;koLFkk dh fLFkfr esa bldh Åijh lrg cQZ ds rqjra vUnj pyh tkrh gSA ?ku ds izkjfEHkd rki dh x.kuk dhft;sA ?ku ,oa cQZ ds ckgj fdlh Hkh Å"ek gkfu dks ux.; eku yhft;sA cQZ dk ?kuRo = 900 fdxzk/eh3 rFkk cQZ ds xyu dh xqIr Å"ek = 3.36 × 105 twy@fdxzk A Ans.

80ºC

Q.7

1kg of ice at 0ºC is mixed with 1 kg of steam at 100ºC. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 105 J/kg and latent heat of vaporization of water = 2.26 × 106 J/kg. 0ºC okyh 1kg cQZ dks 100ºC okyh 1 kg ok"i ds lkFk fefJr fd;k tkrk gSA tc rkih; lkE;koLFkk çkIr gksrh gS rks fudk; dk feJ.k fdl rjg dk gksxkA cQZ ds xyu dh xqIr Å"ek = 3.36 × 105 J/kg vkSj ikuh ds ok"iu dh xqIr Å"ek = 2.26 × 106 J/kg gSA HCV_Ch-25_Ex_7 665 g steam and 1.335 kg water

Ans. Q.8

Ans. Q.9

Ans. Q.10

Calculate the time required to heat 20kg of water from 10ºC to 35ºC using an immersion heater rated 1000W. Assume that 80% of the power input is used to heat the water. Specific heat capacity of water = 4200 J/kgK. 1000 okWV dh be'kZu jkWM dk mi;ksx djds 20 fdxzk ikuh dk rki 10ºC ls 35ºC rd c<+kus esa yxus okys le; dh x.kuk dhft;sA eku yhft;s fd fof'k"V 'kfDr dk 80% gh ikuh xeZ djus gsrq iz;D q r gksrk gSA ikuh dh fof'k"V Å"ek = 4200 twy@fdxzk-K 44 min On a winter day the temperature of the tap water is 20ºC whereas the room temperature is 5ºC. Water is stored in a tank of capacity 0.5 m3 for household use. If it were possible to use the heat liberated by the water to lift a 10 kg mass vertically, how high can it be lifted as the water comes to the room temperature? Take g = 10 m/s2. lnhZ esa fdlh fnu uy ds ikuh dk rki 20ºC gS] tcfd dejs dk rki 5ºC gSA ?kjsyw mi;ksx ds fy;s 0.5 eh3 {kerk okyh Vadh esa ikuh lafpr fd;k tkrk gSA ;fn ikuh ds }kjk eqDr Å"ek dk mi;ksx] 10 fdxzk nzO;eku dks Åij p<+kus esa fd;k tk lds rks ikuh ds dejs ds rki rd vkus ij nzO;eku fdruk Åij mBk;k tk ldrk gSA g = 10 m/s2. k 315 km A bullet of mass 20 g enters into a fixed wooden block with a speed of 40 m/s and stops in it. Find the change in internal energy during the process. 40 eh@ls pky ls xfr djrh gqbZ 20 xzke nzO;eku dh ,d xksyh] ydM+h ds fLFkj xqVds esa izfo"V gks tkrh gSA bl izfØ;k

esa vkarfjd ÅtkZ esa ifjorZu Kkr dhft;sA Ans.

16 J

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Chapter # 25 Calorimetry Q.11 A 50kg man is running at a speed of 18km/h. If all the kinetic energy of the man can be used to increase the temperature of water from 20ºC to 30ºC, how much water can be heated with this energy ? 50 fdxzk nzO;eku okyk ,d O;fDr 18 fdeh@?kaVk dh pky ls nkSM+ jgk gSA ;fn O;fDr dh leLr xfrt ÅtkZ ikuh dk rki 20ºC ls 30ºC rd c<+kus esa iz;qDr dh tk ldrh gks] rks bl ÅtkZ ls fdrus ikuh dk rki c<+k;k tk ldrk gSA Ans. 15 g Q.12

Ans. 13

A brick weighing 4.0kg is dropped into a 1.0 m deep river from a height of 2.0 m. Assuming that 80% of the gravitational potential energy is finally converted into thermal energy, find the thermal energy in calorie. 4.0 fdxzk nzO;eku okyh ,d b±V 2.0 eh- Å¡pkbZ ls 1.0 eh- xgjh unh esa Mky nh tkrh gSA ;fn ;g ekuk tk;s fd xq:Roh; fLFkfrt ÅtkZ dk 80% var esa Å"ek ÅtkZ esa ifjofrZr gks tkrk gS] Å"ek ÅtkZ dk ;g eku dSyksjh esa Kkr dhft;sA 23 cal A van of mass 1500kg travelling at a speed of 54 km/h is stopped in 10s. Assuming that all the mechanical energy lost appears as thermal energy in the brake mechanism, find the average rate of production of thermal energy in cal/s. 54 fdeh@?kaVk pky ls xfr'khy 1500 fdxzk nzO;eku dh ,d osu dks 10 lsd.M esa jksdk tkrk gSA ;g ekurs gq, fd u"V

gksus okyh leLr ;kaf=kd ÅtkZ] czsd O;oLFkk esa rki ds :i esa izdV gksrh gS] Å"ek ÅtkZ ds mRiknu dh vkSlr nj [email protected] esa Kkr dhft;sA Ans.

4000 cal/s

14

A block of mass 100g slides on a rough horizontal surface. If the speed of the block decreases from 10 m/s to 5 m/s, find the thermal energy developed in the process. 100 xzke nzO;eku okyk ,d xqVdk [kqjnjh {kSfrt lrg ij fQlyrk gSA ;fn xqVds dh pky 10 eh@ls ls ?kVdj 5 eh@ls

jg tkrh gS] rks bl izfØ;k esa mRiUu gksus okyh Å"ek Kkr dhft;sA Ans.

3.75 J

15

Two blocks of masses 10kg and 20kg moving at speeds of 10 m/s and 20 m/s respectively in opposite directions, approach each other and collide. If the collision is completely inelastic, find the thermal energy developed in the process. 10 fdxzk rFkk 20 fdxzk nzO;eku okys nks xqVds Øe'k% 10 eh@ls ,oa 20 eh@ls pky ls ijLij foifjr fn'kk esa xfr djrs

gq, vkil esa Vdjk tkrs gSA ;fn VDdj iw.kZr;k vizR;kLFk gS] rks bl izfØ;k esa mRiUu Å"ek Kkr dhft;sA Ans.

3000 J

16.

A ball is dropped on a floor from a height of 2.0 m. After the collision it rises up to a height of 1.5 m. Assume that 40% of the mechanial energy lost goes as thermal energy into the ball. Calculate the rise in the temperature of the ball in the collision. Heat capacity of the ball is 800 J/K. ,d xsna Q'kZ ij 2.0 m. Å¡pkbZ ls fxjkbZ tkrh gSA VDdj ds i'pkr~ ;g iqu% 1.5 eh- Å¡pkbZ rd mB tkrh gSA eku yhft;s fd u"V gksus okyh ;kaf=kd ÅtkZ dk 40% xsna esa ÅtkZ ds :i pyk tkrk gSA VDdj esa xsna ds rki esa o`f) dh x.kuk dhft;sA xsan dh Å"ek /kkfjrk 800 twy@fdxzk gSA 2.5 × 10–3 ºC

Ans. 17.

Ans.

A copper cube of mass 200g slides down on a rough inclined plane of inclination 37º at a constant speed. Assume that any loss in mechanical, energy goes into the copper block as thermal energy. Find the increase in the temperature of the block as it slides down through 60 cm. Specific heat capacity of copper = 420 J/kg-K. 200 xzke nzO;eku okyk rkacs dk ,d ?ku 37º vkufr okys [kqjnjs ur ij fu;r pky ls fQly jgk gSA eku yhft;s fd ;kaf=kd ÅtkZ dh gkfu] rkacs ds ?ku esa Å"ek&ÅtkZ ds :i esa pyh tkrh gSA 60 lseh uhps fQlyus ds i'pkr~ xqVds ds rki esa o`f) Kkr dhft;sA rkacs dh fof'k"V Å"ek = 420 twy@fdxzk-K 8.6 × 10–3 ºC

18.

A metal block of density 6000 kg/m3 and mass 1.2 kg is suspended through a spring of spring constant 200 N/m. The spring-block system is dipped in water kept in a vessel. The water has a mass of 260g and the block is at a height 40cm above the bottom of the vessel. If the support to the spring is broken, what will be the rise in the temperature of the water. Specific heat capacity of the block is 250 J/kg-K and that of water is 4200 J/kg-K. Heat capacities of the vessel and the spring are negligible. 6000 fdxzk0/eh03 ?kuRo okys rFkk 1.2 fdxzk0 nzO;eku okys /kkrq ds xqVds dks 200 U;wVu/eh0 okys fLçax fu;rkad ls yVdk;k tkrk gSA fLçax&xqVds fudk; dks ikuh ls Hkjs ik=k esa Mqcks;k tkrk gSA ikuh dk nzO;eku 260 g gS rFkk xqVdk ik=k ds isna s ls 40 lseh0 Å¡pkbZ ij gSA vxj fLçax ftl ij c¡/kk gS og lgkjk VwV tkrk gS rks ikuh ds rkieku esa fdruh o`f) gksxhA xqVds dh fof'k"V Å"ek 250 J/kg-K rFkk ikuh dh 4200 J/kg-K gSA ik=k rFkk fLçax dh fof'k"V Å"ek dks ux.; ekfu,A

Ans.

0.003ºC

HCV_Ch-25_Ex_18

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