Chapter
11
Thermal Properties of Matter Solutions SECTION - A School/Board Exam. Type Questions Very Short Answer Type Questions : 1.
You put a bottle of soft drink in a refrigerator and leave it unitil its temperature dropped by 10.0 K. What is its temperature change in C°.
Sol. 10 C° 2.
Why are we advised to store medicine below 86°F?
Sol. A temperature of 86 °F is approximately equal to 30°C. Above this temperature, chemical reactions may take place resulting in a change in the composition of the medicine. The medicine may then become ineffective or even harmful. 3.
Why does a blackbody appear brighter than the polished surface when both are heated to the same temperature?
Sol. This is due to the reason that compared to a polished surface, a blackbody is a good emitter. 4.
Why is it hotter at the same distance over the top of a fire than in front of it?
Sol. In front of the fire, heat is received only by radiation. But at the same distance over the fire, heat is received both due to radiation as well as convection. 5.
A beaker full of hot water is placed on a wooden table. Does it lose heat? If yes, in what ways?
Sol. The beaker lose heat mainly due to radiation. A small part of heat will be lost to the wooden table on account of conduction also. 6.
It is warmer to wear two thin shirts than a single thick shirt of the same material. Why?
Sol. It is on account of the reason that air, which is very poor conductor of heat, is enclosed between two thin blankets (or shirts). Thus, the heat loss from the body to the surroundings is less than when we use a single thick blanket (or shirts) 7.
What is the S.I. unit of heat energy?
Sol. joule (J) Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
60 8.
Thermal Properties of Matter
Solutions of Assignment (Set-1)
Define the term Triple point for water.
Sol. Triple point of water is the temperature at which the three phases of water (ice, water, water vapour) are all equally stable and coexist in equilibrium. 9.
What is the dimensional formulae of coefficient of linear expansion ().
Sol. [M0L0T0K–1] 10. What do you mean by thermal stress? Sol. Thermal stress is equal to force per unit area developed when thermal expansion or contraction is resisted. Short Answer Type Questions : 11. Define the water equivalent of a body briefly. Sol. Water Equivalent : If the heat capacity of a body is expressed in term of mass of water, it is called water equivalent of the body. Water equivalent of a body is the quantity of water whose temperature would be raised through 1 C° (or 1K) by the same amount of heat as required to raise the temperature of the body through 1 C° (or 1 K). It is denoted by w and is numerically equal to the product of mass (m) of the body and its specific heat. 12. Distinguish melting and freezing points. Sol. The temperature at which a solid changes into a liquid is called melting point and the temperature at which a liquid changes into a solid is called its freezing point. For most crystalline substances, the melting and the freezing points are the same. 13. What is the dimensional formulae of thermal conductivity of a material? Qx , K Sol. As K = A T1 – T2 t
⎡⎣ML2T –2 ⎤⎦ L = [MLT–3K–1] ⎡⎣L2 ⎤⎦ .K T
14. State the Newton’s law of cooling. Sol. Newton’s law of Cooling : It is a common observation that bodies at temperatures higher than that of the surroundings lose heat. Newton was the first to make a scientific study in this direction. From experimental observations, he came to a conclusion which is known as the Newton’s law of cooling according to which : The rate of loss of heat of a body is directly proportional to the excess of the temperature of the body w.r.t., the surroundings –
d ( – 0 ) . dt
15. What do you mean by thermal capacity of a body? Sol. Heat capacity or Thermal capacity of a body, if instead of unit mass, we consider the whole body, we talk of its heat capacity or thermal capacity. It is defined as the amount of heat required to raise the temperature of the body by one degree. 16. Two beakers of water, A and B, initially are at the same temperature. The temperature of the water in beaker A is increased by 10 F°, and the temperature of the water in beaker B is increased by 10 K. After these temperature changes, which beaker of water has the higher temperature? Explain. Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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61
Sol. Rise in temperature in beaker A = 10Fº
=
5 ⎡ ⎤ ⎢∵ k 9 F ⎥ ⎣ ⎦
5 [10]K 9
< 10 K Beaker B is at higher temperature. 17. If an area measured on the surface of a solid body is A0 at some initial temperature and then changes by A when the temperature changes by T, show that A = 2 A0 T, where is the coefficient of linear expansion. Sol. A = A0(1 + T) = coefficient of area expansion we know that = 2 A = A0[1 + 2T] A = A0 + 2A0T [A = 2A0T] 18. A circular sheet of aluminium is 55.0 cm in diameter at 15.0°C. By how much does the area of one side of the sheet change when the temperature increases to 27.5°C? Sol. A = (2)A0.T 19. You are given a sample of metal and asked to determine its specific heat. You weigh the sample and find that its weight is 28.4 N. You carefully add 1.25 × 104 J of heat energy to the sample and find that its temperature rises by 18Cº? What is the sample’s specific heat? Sol. Q = mcT, Given weight = 28.4 N, Q = 1.25 × 104 J, T = 18.0C°
mg = 28.4 N
1.25 × 104 =
28.4 c.18 9.8
c=
1.25 10 4 9.8 28.4 18
c=
12.25 10 4 511.2
c = 2.4 × 102 J/kgC° 20. Show that the SI units of thermal conductivity is W/m K.
Sol. H =
W H.L W.m KA .T , K = , hence S.I. unit of K is = 2 = mK A.T L m .K
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Thermal Properties of Matter
Solutions of Assignment (Set-1)
21. You are asked to design a cylindrical steel rod 50.0 cm long with a circular cross-section, that will conduct 150.0 J/s from a furnace at 400°C to a container of boiling water under 1 atmosphere. What must the rod’s diameter be ? (Ksteel = 50.2 W/m K).
Sol.
Q KA T1 – T2 H 4L 150 4 0.5 K d 2 2 10 4 cm2 H T d t L K T 50.2 300 4L d = 7.6 cm
22. If the length of a cylinder on heating increases by 2%, calculate the increment of its base area. Sol. If length increases by 2% on heating, radius will also increase by 2%. As base area (radius)2, so it will increase by 4% 23. The coefficient of volume expansion of a liquid is 49 × 10–5/K. Calculate the fractional change in its density when the temperature is raised by 30°C. Sol. When the temperature of a liquid increases by T C°, the mass will remain unchanged while due to thermal expansion volume will increase and become V = V(1 + T ), where is coefficient of volume expansion of liquid.
=
=
m V m V 1 T 1 T
T ⎡ ⎤ Fractional change in density = ⎢1 – ⎥ = 1 T ⎦ ⎣
Fractional change in density =
49 10 –5 30 1.5 × 10–2 1 49 10 –5 30
24. One g of ice at 0°C is mixed with one g of steam at 100°C. After thermal equilibrium is reached, what is the temperature of mixture? Sol. Heat required by ice to convert into water at 100°C, Q1 = 1 × 80 + 1 × 1 × 100 = 180 cal Heat supplied by steam if it was to condense totally and convert into water at 100°C. Q2 = 1 × 540 = 540 cal As Q2 > Q1, entire steam will not condense and final temperature = 100°C 25. A lead ball moving with velocity v strikes a wall and stops. If 50% of its energy is converted into heat, then what will be the increase in temperature? (Specific heat of Lead is s) [J = mechanical equivalent of heat]
1 mv 2 v2 50 2 = H = msT or T = Sol. 4 Js 100 J Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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63
26. Two rods of same length and material transfer a given amount of heat in 12 s, when they are joined end to end. But when they are joined lengthwise, find time taken to transfer the same amount of heat in same condition. Sol. Q =
KA T1 – T2 2l
t1
K (2 A)(T1 – T2 ) t t2 t2 1 3 s 4 l
27. A body initially at 80°C cools to 64°C in 5 min and from 64°C to 52°C in 10 min. Calculate the temperature of the surrounding. Sol. According to Newton’s law of cooling In the First case,
80 – 64 ⎡ 80 64 ⎤ – TS ⎥ K ⎢ 5 2 ⎣ ⎦
3.2 = K [72 – TS ]
In the Second case,
...(i)
⎡ 64 52 ⎤ 64 – 52 – TS ⎥ K ⎢ 5 ⎣ 2 ⎦
2.4 = K [58 – TS ]
...(ii)
From equations (i) & (ii), TS = 16°C 28. A block of mass 1000 g 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. Sol. Thermal energy = Change in kinetic energy Thermal energy = 37.5 J 29. Show that moment of inertia of a solid body of any shape changes with temperature as I = I0 (1 + 2), where I0 is the moment of inertia at 0°C and is the coefficient of linear expansion of solid. Sol. I0 = k mr02
I = Moment of inertia m = Mass of the solid r0 = Distance of a point from axis of rotation
If change in temperature is , then new distance becomes r = r0(1 + )
I = kmr2 = kmr02(1 + )2 = kmr02(1 + 2)
I = l0 (1 + 2 )
30. A bullet of mass 200 mg enters a fixed block with speed 50 m/s and stops in it. Find the change in internal energy during the process. Sol. Change in internal energy = Change in kinetic energy =
1 2 200 10 –3 50 2
= 2500 × 100 × 10–3 = 250 J Change in internal energy 250 J Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Thermal Properties of Matter
Solutions of Assignment (Set-1)
Long Answer Type Questions : 31. Derive Newton’s law of cooling. Can you apply Newton’s law of cooling to the case of an object which is slightly cooler than its surroundings? Sol. As per the statement of the law, – or –
dQ = K(T – T0) dt
dQ (T – T0) dt ...(i)
Here, K is a constant of proportionality which is positive and its value depends upon the area and nature of the radiating surface.
As dQ = mc dT,
dQ dT mc dt dt
...(ii)
From equations (i) and (ii), – mc.
or
dT K T – T0 dt
K dT dt = – kdt = – mc T – T0
where k =
...(iii)
K = a constant mc
Integrating both sides of equation (iii), we get dT
∫T – T
0
= – k ∫ dt
or ln (T – T0) = – kt + c
...(iv)
No, we connot apply Newton’s law of cooling if temperature is below than surroundings. 32. Derive the expression for the rate of flow of heat energy through a conductor maintained at different temperature at its two ends. Sol. Heat transfer occurs only between regions that are at different temperatures, and the direction of heat flow is always from higher to lower temperature. Figure given below shows a rod of conducting material with crosssectional area A and length L. The left end of the rod is kept at a temperature T1 and the right end at a lower temperature T2, so heat flows from left to right. The sides of the rod are covered by an insulator, so no heat transfer occurs at the sides.
T1
A
T2
L Fig. (a) Steady state heat flow due to conduction in a uniform rod Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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65
When a quantity of heat dQ is transferred through the rod in a time dt, the rate of heat now is dQ/dt. We dQ call this rate the heat current, denoted by H. That is, H = . Experiments show that the heat current is dt proportional to the cross-sectional area A of the rod and to the temperature difference (T1 – T2) and is inversely proportional to the rod length L. Introducing a proportionality constant K called the thermal conductivity of the material, we have H=
dQ KA T – T2 dt L 1
33. Define the coefficient of thermal conductivity and explain the formula used. Why do metals have higher conductivity than insulator? Sol. Coefficient of thermal conductivity of a substance is numerically equal to heat conducted during steady state in unit time through unit area of any section of the substance under unit temperature gradient (the heat flow being normal to the area). Formula used Q KA T1 – T2 t x
Where,
Q Rate of flow of heat t
A Cross-sectional area T1 – T2 temperature gradient x
K coefficient of thermal conductivity Thermal conduction requires a material medium. Different substances have different abilities to conduct heat. Iron conducts heat easily than glass. This fact is expressed by saying that the thermal conductivity of iron is higher than that of glass. Metals are the best conductors of heat. Liquids are poor conductors and all liquids have roughly the same thermal conductivity. The conductivity of gases is much less than that of liquids. Bad conductors are also called non conductors or insulators. 34. How Newton’s law of cooling can be experimentally verified? Sol. Experimental arrangement : The apparatus used is shown in figure. It consists of a copper calorimeter (C) blackened from outside (so as to ensure the same nature of its entire radiating surface) and fitted with a cork containing a half degree thermometer (T ), a stirrer (s). Water heated to about 30°C to 40°C above the room temperature is put in the calorimeter. It is then suspended in a double walled vessel V. The space between the wall of this vessel is filled with water at room temperature. This serves as a constant temperature enclosure. Another half degree thermometer (T0) is used to measure the temperature of the enclosure.
S
T0
log (T – T0)
T
C V (a)
(b)
t
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Thermal Properties of Matter
Solutions of Assignment (Set-1)
Water is gently stirred and its temperature (T) is noted regularly first after every half minute, then after every minute and towards the end after every two minutes. The process is repeated till the temperature of water in the calorimeter is about 5°C above the mean temperature (T0) of the enclosure. The values of (T – T0) are calculated for various values of time t. A graph is then plotted between ln(T – T0) and t. The graph is found to be a straight line as shown in Figure (b), thereby verifying Newton’s law of cooling. 35. 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 Wm–1°C–1 and that of steel = 46 Wm–1°C–1 0°C
Copper
Steel
100°C
Sol. KCu = 390 W/m°C Now, since they are in series connection, so, the heat current is the same. So, H1 = H2 K Cu .A T – 0 l
=
K s .A 100 – T l
0°C
Copper
Steel
100°C
T
T = 10.55 °C 36. A composite slab is prepared by pasting two plates of thickness l1 and l2 and thermal conductivities K1 and K2. The slabs have equal cross-sectional area. Find the equivalent conductivity of the composite slab. Sol. Here the thermal resistors are in series Rseries = R1 + R2
Q,T1 l1 l2
1 (l1 l 2 ) 1 l1 1 l2 K A K1 A K 2 A
Q,T2
K
K1K 2 (l1 l 2 ) K1l 2 K 2 l1
37. 1000 g of ice at 0°C is mixed with 1000 g of steam at 100°C. What will be the composition of the system when thermal equilibrium is reached ? Latent heat of fusion of ice = 3.36 × 105 J kg–1 and latent heat of vaporisation of water = 2.26 × 106 J/kg. Sol. Lfusion = 3.36 × 105 J kg–1 Lvap = 2.26 × 106 J kg–1 Heat required by 1000 g i.e. 1 kg of ice to melt, Q1 = 3.36 × 105 J. Heat required by 1 kg water at 0°C to attain 100°C Q1 = 4186 × 100 = 4.2 × 105 J Heat released when 1000 g i.e. 1 kg of steam condenses, Q2 = 2.26 × 106 J
∵
Q1 + Q1 = 7.56 × 105J < Q2
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Thermal Properties of Matter
67
Whole of ice will melt and heat upto 100°C. Amount of steam condensed in the process, 7.56 105 m = 2.26 106
= 0.334 kg = 334 g Final composition of the mixture is 1334 g of water and 666 g of steam. 38. The temperatures of equal masses of three different liquids a, b and c are 12°C, 19°C and 28°C respectively. The equilibrium temperature when a and b are mixed is 16°C, and when b and c are mixed, it is 23°C. What will be the temperature when a and c are mixed? Sol. The temperature of a = 12°C The temperature of c = 28°C The temperature of b + c = 23° The temperature of b = 19°C The temperature of a + b = 16° when a & b are mixed, Mca (16 – 12) = Mcb (19 – 16)
...(i)
when b and c are mixed Mcb (23 – 19) = Mcc(28 – 23)
...(ii)
When a & c are mixed, if T is the common temperature of mixture. Mca(T – 12) = Mcc(28 – T)
...(iii)
From equations (i), (ii) and (iii)
T=
628 20.3C 31
39. If a thermometer reads freezing point of water as 20°C and boiling point as 150°C, find the reading of thermometer when the actual temperature is 60°C.
Sol.
100 – 60 150 – x x – 20 60 – 0 or, 40(x – 20) = 60 (150 – x) x = 98°C
40. On a new scale of temperature (which is linear) and called the w scale, the freezing and boiling points of water are 39°w and 239°w respectively. Calculate the temperature on the new scale, corresponding to a temperature of 39°C on the Celsius scale. Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Thermal Properties of Matter
New scale
Sol. 100°C
Solutions of Assignment (Set-1)
239°w
100 division 0°C
39°w
Temperature on new scale
= 39 × 2 + 39 = (78 + 39)°w = 117°w
41. A crystal has a coefficient of expansion 13 × 10–7 per °C in one direction and 231 × 10–7 per °C in every other direction at right angles to it. Calculate the volume expansion of the crystal. Sol. = 1 + 2 + 3 = 13 × 10–7 + 231 × 10–7 + 231 × 10–7 = 475 × 10–7 per °C 42. When the temperature of a rod increases from t to t + t, its moment of inertia increases from I to I + I. If be the coefficient of linear expansion of the rod, then find the value of
I . I
Sol. Moment of inertia of rod I=
1 ml 2 12
I =
1 2ml 2 .l 12
...(i)
...(ii)
Divide equation (ii) by equation (i), we get l L 2. I L
As L = L T or
L T L
l 2T l
43. A tap supplies water at 10°C and another tap at 100°C. Calculate the amount of hot water required to get 20 kg water at 35°C. Sol. m(100 – 35) × C = (20 – m) × (35 – 10) × C 65m = 500 – 25m 90m = 500
m=
500 5.6 kg 90
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69
44. Two vessels of different materials are similar in size in every respect. The same quantity of ice filled in them gets melted in 20 minutes and 40 minutes respectively. What is the ratio of thermal conductivities of the metals? Sol.
Q K t
Kt = Q
i.e., Kt = constant
K1 t2 K2 t1
K1 2 = 1 K2
45. The specific heat of many solids at low temperature varies with absolute temperature T according to the relation s = aT3, where a is constant. What is the heat energy required to raise the temperature of a mass m of such a solid from T = 0 K to T = 20 K? 20
Sol. Q =
∫ ms dT 0
20
=
∫ m aT dT 3
0
20
⎡T 4 ⎤ = ma ⎢ ⎥ ⎣ 4 ⎦0
Q = 4 × 104 ma J
SECTION - B Model Test Paper Very Short Answer Type Questions : 1.
Is it necessary that every black coloured object should be considered a blackbody?
Sol. No 2.
To raise the temperature of an object, must you add heat to it? If you add heat to an object, must you raise its temperature?
Sol. No. During phase transformation no raise in temperature. 3.
Do water and ice have the same specific heats?
Sol. No, water and ice do not have the same specific heat. The specific heat of water is 1 cal/gCº and that of ice is 0.5 cal/gC°. Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Thermal Properties of Matter
Solutions of Assignment (Set-1)
Why does the blowing of hydrogen through electrical machines cool them more than when the same mass of air is used for this purpose?
Sol. The reason for this is that the thermal conductivity of hydrogen is more than that of air. 5.
Write the S.I. unit of thermal stress.
Sol. N/m2. 6.
At what common temperature, would a block of wood and a block of metal feel equally cold or equally hot when touched?
Sol. When the temperature of the object (both wood and metal) is equal to that of human body. 7.
Pieces of copper and glass are heated to the same temperature. Why does the piece of copper feel hotter on touching?
Sol. We know that copper is a far better conductor of heat than glass. When we touch the hot copper piece, it transmits heat readily to the hand. But this is not the case when the hot glass piece is touched. 8.
Why does a glass dish shatter when taken from the oven and put into cold water?
Sol. The dish is shattered by large stresses due to non-uniform contraction of glass. Short Answer Type Questions : 9.
Why do electrons in insulator not contribute to its conductivity?
Sol. In insulators, there are no free electrons. Only free electrons conduct heat and as such electrons do not contribute towards the conductivity of insulators. 10. You feel sick and are told that you have a temperature of 40.2°C. What is your temperature in °F? Should you be concerned?
Sol. TF =
9 TC 32 5
TC = 40.2
TF =
9 40.2 32 5
TF =104.9°F This temperature is greater than normal temperature of the body, hence you have to be concerned. 11. A copper block of mass 100 g is heated till its temperature is increased by 20°C. Find the heat supplied to the block. Specific heat capacity of copper = 0.09 cal/g°C Sol. The heat supplied is Q = mST = 100 × 0.09 × 20 = 180 cal Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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12. Find the thermal resistance of an aluminium rod of length 10 cm and area of cross-section 2 cm2. The heat current is along the length of rod. Thermal conductivity of aluminium = 200 Wm–1K–1. Sol. The thermal resistance is R =
x kA
10 10 –2 m R= 200 Wm–1K –1 2 10 –4 m2 R = 2.5 KW–1 13. Express a temperature of 60°F in degree celsius and in kelvin. Sol. As we know that TC =
5 TF – 32 9
5 60 – 32 15.55C 9
TC = Also,
T = TC + 273.15 = 15.55°C + 273.15 T = 288.70 K 14. The scale on a steel metre stick is calibrated at 15°C. What is the error in the reading of a length of 60 cm at 27°C? steel = 1.2 × 10–5 0C–1. Sol. At higher temperatures actual reading is more than the scale reading. The error in the reading will be l = (scale reading) () (T) = (60) (1.2 × 10–5) (27 – 15) = 0.00864 cm 15. The steam point and the ice point of a mercury thermometer are marked as 80° and 10°. At what temperature on centigrade scale will the reading of this thermometer be 59°? Sol.
C–0 59 – 10 49 100 – 0 80 – 10 70
c 70º C
Short Answer Type Questions : 16. What do you mean by latent heat? Explain. Sol. The heat necessary to change a unit mass of a substance from one phase to another at the same temperature and pressure is called its latent heat (L). Thus, the amount of heat required for melting and vaporizing a substance of mass m are given as, Q = mL For a solid-liquid transition, the latent heat is known as the latent heat of fusion (Lf) and for liquid–gas transition, it is known as the latent heat of vaporization (LV) Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Solutions of Assignment (Set-1)
17. How much heat is required to convert 8.0 g of ice at – 15°C to steam at 100°C? [Given, Cice = 0.53 cal/g°C, Lf = 80 cal/g and LV = 539 cal/g and Cwater = 1 cal/g°C]
Ice
Ice
Sol. – 15°C
Q1
0°C
Water Q2
0°C
Water Q3
100°C
Steam Q4
100°C
Q1 = mCice(Tf – Ti) = (8.0) (0.53) (0 + 15) = 63.6 cal Q2 = mLf = (8.0) (80) = 640 cal Q3 = mCwater (Tf – Ti) = (8.0) (1.0) (100 – 0) = 800 cal Q4 = mLV = (8.0) (539) = 4312 cal
Net heat required
Q = Q1 + Q2 + Q3 + Q4 = 5815.6 cal 18. A glass beaker holds exactly 1 litre at 0°C. What is its volume at 50°C? [Given, g = 8.3 × 10–6 0C–1] Sol. The volume of beaker after the temperature change is, Vbeaker = V0(1 + 3 g.T) = (1) [1 + 3 × 8.3 × 10–6 × 50] = 1.001 litre
Vbeaker = 1.001 litre
19. Obtain the expression for the force developed in a rod which is under the thermal stress. Sol. When a rod whose end are rigidly fixed (such as to prevent from expansion or contraction) undergoes a change in temperature, thermal stresses are developed in the rod. This is because if the temperature is increased, the rod has a tendency to expand but since, it is fixed at two ends, the rod exerts a force on supports.
Thermal strain =
l T l
l,
So, thermal stress = (Y) (thermal strain) = YT or, force on supports F = A (stress) = YAT Here, Y = Young’s modulus of elasticity of the rod. F = YAT
20. Show that the volume thermal expansion coefficient for an ideal gas at constant pressure is
1 . T
Sol. For an ideal gas PV = nRT As P is constant, we have PdV = nRdT Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Thermal Properties of Matter
dV nR dT P
or,
=
1 dV nR nR 1 . V dT PV nRT T
=
1 T
73
21. Suppose a liquid in a container is heated at the top rather than at the bottom. What is the main process by which the rest of the liquid becomes hot? Sol. The main process by which the rest of the liquid becomes hot is conduction. This is because in convection flow of liquid is required and this does not happen in this situation. Long Answer Type Questions : 22. A copper rod 2 m long has a circular cross section of radius 1 cm. One end is kept at 100°C and the other at 0°C, and the side surface is insulated so that negligible heat is lost through the surface, find : (i)
The thermal resistance of the bar
(ii)
The thermal current (H)
⎡ dT ⎤ (iii) The temperature gradient ⎢ ⎥ and ⎣ dx ⎦ (iv)
The temperature 25 cm from the hot end. (Thermal conductivity of copper is 401 Wm–1K–1)
Sol. (i)
Thermal resistance, R =
R=
(ii)
2
401 10–2
2
Thermal current, H =
l l kA k .r 2
= 15.9 KW–1
T 100 R 15.9
H = 6.3 W
(iii) Temperature gradient = (iv)
0 – 100 – 50 Km–1 50 ºC(m–1 ) 2
Let T1 be the temperature at 25 cm from the hot end, then (T1 – 100) = (temperature gradient) × (distance) T1 – 100 = (– 50) × (0.25)
100°C
T1
0°C
0.25 m 2.0 m
T1 = 87.5 °C Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
74
Thermal Properties of Matter
Solutions of Assignment (Set-1)
23. A hot body placed in air is cooled according to Newton’s law of cooling, the rate of decrease of temperature being K times the temperature difference from the surroundings. Starting from t = 0, find the time in which the body will lose half the maximum temperature it can lose. Sol. Let T0 be the temperature of surrounding and T the temperature of hot body at some instant. Then –
dT k T – T0 dt T
or
t
dT ∫T T – T0 – k ∫0 dt m
Tm = temperature at time t = 0
Solving this equation, we get T = T0 + (Tm – T0) e–kt
T – T0 = (Tm – T0)e–kt
Given that,
T – T0 =
Tm – T0 2
Tm – T0 Tm – T0 e – kt 2
t=
ln 2 k
Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
