Q: A deep rectangular pond of surface area A, containing water (density = \rho, specific heat capacity = s), is located in a region where the outside air temperature is a steady value at the -26^\circC. The thickness of the frozen ice layer in this pond, at a certain instant is x. Taking the thermal conductivity of ice as K, and its specific latent heat of fusion as L, the rate of increase of the thickness of ice layer, at this instant would be given by

26K/(\rho xL). If the cross-section is not uniform or steady state is not reached, the heat conduction equation applies to a thin layer perpendicular to the heat flow. Rate of heat flow for ice growth: \dfrac{d\theta}{dt} = \dfrac{KA(\theta_0 - \theta_1)}{x}, where d\theta = \rho A\,dx\,L, \theta_0 = 0^\circC and \theta_1 = -26^\circC. Thus \dfrac{\rho A\,dx\,L}{dt} = \dfrac{KA(\theta_0 - \theta_1)}{x} \Rightarrow \dfrac{dx}{dt} = \dfrac{K[0-(-26)]}{\rho xL} = \dfrac{26K}{\rho xL}.

Question 1

A cup of coffee cools from 90^\circC to 80^\circC in t minutes, when the room temperature is 20^\circC. The time taken by a similar cup of coffee to cool from 80^\circC to 60^\circC at a room temperature same at 20^\circC, is

Accepted Answer

\dfrac{13}{5}t. By Newton's law of cooling, \dfrac{\Delta T}{t}=K\left[\dfrac{T_i+T_f}{2}-T_0ight]. First case: \dfrac{90-80}{t}=K\left[\dfrac{90+80}{2}-20ight]\Rightarrow\dfrac{10}{t}=65K\Rightarrow K=\dfrac{2}{13t}. Second case: \dfrac{80-60}{t_1}=\dfrac{2}{13t}\left[\dfrac{80+60}{2}-20ight]=\dfrac{2}{13t}	imes50\Rightarrow t_1=\dfrac{13}{5}t.

Question 2

Three stars A, B, C have surface temperatures T_A, T_B, T_C respectively. Star A appears bluish, star B appears reddish and star C yellowish. Hence

Accepted Answer

T_A > T_C > T_B. According to Wien's displacement law, \lambda = \dfrac{b}{T}, i.e. \lambda \propto \dfrac{1}{T}. We know that \lambda_{bluish} < \lambda_{yellowish} < \lambda_{reddish}. Hence, T_A > T_C > T_B.

Question 3

An object kept in a large room having air temperature of 25^\circC takes 12 minutes to cool from 80^\circC to 70^\circC. The time taken to cool for the same object from 70^\circC to 60^\circC would be nearly

Accepted Answer

15 min. From Newton's law of cooling, \dfrac{T_1 - T_2}{t} = \dfrac{1}{K}\left(\dfrac{T_1 + T_2}{2} - T_0ight) with T_0 = 25^\circC. Cooling from 80^\circC to 70^\circC in 12 min: \dfrac{80-70}{12} = \dfrac{1}{K}\left(\dfrac{80+70}{2} - 25ight) \Rightarrow \dfrac{5}{6} = \dfrac{50}{K} ...(i). Cooling from 70^\circC to 60^\circC: \dfrac{70-60}{t} = \dfrac{1}{K}\left(\dfrac{70+60}{2} - 25ight) \Rightarrow \dfrac{10}{t} = \dfrac{40}{K} ...(ii). Dividing (i) by (ii): \dfrac{5}{6} 	imes \dfrac{t}{10} = \dfrac{50}{40} \Rightarrow t = \dfrac{5}{4} 	imes 12 = 15 minutes.

Question 4

A deep rectangular pond of surface area A, containing water (density = \rho, specific heat capacity = s), is located in a region where the outside air temperature is a steady value at the -26^\circC. The thickness of the frozen ice layer in this pond, at a certain instant is x. Taking the thermal conductivity of ice as K, and its specific latent heat of fusion as L, the rate of increase of the thickness of ice layer, at this instant would be given by

Accepted Answer

26K/(ho xL). If the cross-section is not uniform or steady state is not reached, the heat conduction equation applies to a thin layer perpendicular to the heat flow. Rate of heat flow for ice growth: \dfrac{d	heta}{dt} = \dfrac{KA(	heta_0 - 	heta_1)}{x}, where d	heta = ho A\,dx\,L, 	heta_0 = 0^\circC and 	heta_1 = -26^\circC. Thus \dfrac{ho A\,dx\,L}{dt} = \dfrac{KA(	heta_0 - 	heta_1)}{x} \Rightarrow \dfrac{dx}{dt} = \dfrac{K[0-(-26)]}{ho xL} = \dfrac{26K}{ho xL}.

Question 5

The power radiated by a black body is P and it radiates maximum energy at wavelength \lambda_0. If the temperature of the black body is now changed, so that it radiates maximum energy at wavelength \dfrac{3}{4}\lambda_0, the power radiated by it becomes nP. The value of n is

Accepted Answer

\dfrac{256}{81}. By Wien's law, \lambda_{max} T =  constant, so \dfrac{T_1}{T_2} = \dfrac{\lambda_{max_2}}{\lambda_{max_1}}. With \lambda_{max_1} = \lambda_0 and \lambda_{max_2} = \dfrac{3}{4}\lambda_0, \dfrac{T_1}{T_2} = \dfrac{3}{4}. By Stefan's law P \propto T^4, so \dfrac{P_1}{P_2} = \left(\dfrac{T_1}{T_2}\right)^4 = \left(\dfrac{3}{4}\right)^4 = \dfrac{81}{256}. With P_1 = P and P_2 = nP: \dfrac{P}{nP} = \dfrac{81}{256} \Rightarrow n = \dfrac{256}{81}.

Question 6

Two rods A and B of different materials are welded together as shown in figure. Their thermal conductivities are K_1 and K_2. The thermal conductivity of the composite rod will be

Accepted Answer

\dfrac{K_1 + K_2}{2}. The rods are in parallel with equal cross-sectional areas and the same length, so the composite conductivity is K = \dfrac{K_1 A + K_2 A}{2A} = \dfrac{K_1 + K_2}{2}.

Question 7

A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be

Accepted Answer

1800. Radiated power of a black body, P=\sigma A T^4, where A is the surface area and T the temperature. When the radius is halved, the new area A'=\dfrac{A}{4}. So P'=\sigma\left(\dfrac{A}{4}ight)(2T)^4=\dfrac{16}{4}\,\sigma A T^4=4P=4	imes450=1800 watts.

Question 8

A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U_1, at wavelength 500 nm is U_2 and that at 1000 nm is U_3. Wien's constant, b=2.88	imes10^6 nmK. Which of the following is correct?

Accepted Answer

U_2>U_1. According to Wien's law, \lambda_m T=b, so \lambda_m=\dfrac{b}{T}=\dfrac{2.88	imes10^6	ext{ nmK}}{5760	ext{ K}}=500 nm. Since \lambda_m=500 nm is the wavelength corresponding to maximum energy, U_2>U_1.

Question 9

A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be

Accepted Answer

\dfrac{3}{2}T. By Newton's law of cooling, \Delta T=\Delta T_0\,e^{-\lambda t}. For the first 10 min, 3T-2T=(3T-T)e^{-\lambda	imes10}, giving e^{-10\lambda}=\dfrac{1}{2}. Measuring the excess temperature over the full 20 min from the start, T'-T=(3T-T)e^{-\lambda	imes20}=2T\,(e^{-10\lambda})^2=2T\left(\dfrac{1}{2}ight)^2=\dfrac{T}{2}. Hence T'=T+\dfrac{T}{2}=\dfrac{3T}{2}.

Question 10

The two ends of a metal rod are maintained at temperatures 100^\circC and 110^\circC. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200^\circC and 210^\circC, the rate of heat flow will be

Accepted Answer

4.0 J/s. Here \Delta T_1=110-100=10^\circC with \dfrac{dQ_1}{dt}=4 J/s, and \Delta T_2=210-200=10^\circC. The rate of heat flow is directly proportional to the temperature difference, which is the same (10^\circC) in both cases. Hence \dfrac{dQ_2}{dt}=4 J/s.

Question 11

Certain quantity of water cools from 70^\circC to 60^\circC in the first 5 min and to 54^\circC in the next 5 min. The temperature of the surroundings is

Accepted Answer

45^\circC. Apply Newton's law of cooling with surrounding temperature t^\circC. First case: \dfrac{70-60}{5}=K(65-t), where 65^\circC is the average of 70 and 60. Second case: \dfrac{60-54}{5}=K(57-t), where 57^\circC is the average of 60 and 54. Dividing, \dfrac{10}{6}=\dfrac{65-t}{57-t}, which gives t=45^\circC.

Question 12

A piece of iron is heated in a flame. If first becomes dull red then becomes reddish yellow and finally turns to white hot. The correct explanation for the above observation is possible by using

Accepted Answer

Wien's displacement law. As the temperature rises, the wavelength of maximum emission decreases (red 	o yellow 	o white). This is explained by Wien's displacement law, \lambda_m T= constant.

Question 13

If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q? (\sigma stands for Stefan's constant.)

Accepted Answer

(Q/4\pi R^2\sigma)^{1/4}. From Stefan's law, E=\sigma T^4. The rate of energy production Q=E	imes A=\sigma T^4	imes4\pi R^2. Hence the temperature of the star T=\left(\dfrac{Q}{4\pi R^2\sigma}ight)^{1/4}.

Question 14

A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod when placed in thermal contact with the two reservoirs in time t?

Accepted Answer

Q/16. In steady state Q=\dfrac{KA(	heta_1-	heta_2)t}{l}, so \dfrac{Q}{t}\propto\dfrac{A}{l}\propto\dfrac{r^2}{l}. On remelting into a rod of half the radius, volume is conserved: \pi r_1^2 l_1=\pi r_2^2 l_2 with r_2=\dfrac{r_1}{2}, giving l_2=4l_1. Then \dfrac{Q_1}{Q_2}=\dfrac{r_1^2}{l_1}	imes\dfrac{l_2}{r_2^2}=\dfrac{r_1^2}{l_1}	imes\dfrac{4l_1}{(r_1/2)^2}=16. Hence Q_2=\dfrac{Q_1}{16}=\dfrac{Q}{16}.

Question 15

A black body at 227^\circC radiates heat at the rate of 7 cal cm^{-2} s^{-1}. At a temperature of 727^\circC, the rate of heat radiated in the same units will be

Accepted Answer

112. By Stefan's law E=\sigma T^4. With T_1=227+273=500 K and T_2=727+273=1000 K, \dfrac{E_1}{E_2}=\left(\dfrac{T_1}{T_2}\right)^4\Rightarrow E_2=7\left(\dfrac{273+727}{273+227}\right)^4=7\left(\dfrac{1000}{500}\right)^4=112 cal cm^{-2} s^{-1}.

Question 16

The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T_1 and T_2 (T_1>T_2). The rate of heat transfer, \dfrac{dQ}{dt}, through the rod in a steady state is given by

Accepted Answer

\dfrac{dQ}{dt}=\dfrac{KA(T_1-T_2)}{L}. For a rod of length L and cross-sectional area A with faces maintained at temperatures T_1 and T_2, in steady state the rate of heat flow from one face to the other is \dfrac{dQ}{dt}=\dfrac{KA(T_1-T_2)}{L}.

Question 17

A black body is at 727^\circC. It emits energy at a rate which is proportional to

Accepted Answer

(1000)^4. By Stefan's law, E\propto T^4 (or E=\sigma T^4). Converting to kelvin, T=727+273=1000 K, so E\propto(727+273)^4=(1000)^4.

Question 18

Assuming the sun to have a spherical outer surface of radius r, radiating like a black body at temperature t^\circC, the power received by a unit surface, (normal to the incident rays) at a distance R from the centre of the sun is (where \sigma is the Stefan's constant)

Accepted Answer

\dfrac{r^2\sigma(t+273)^4}{R^2}. By Stefan's law, the power radiated at the sun's surface is P=\sigma	imes4\pi r^2 T^4 (with e=1 for a black body). The intensity received at a distance R\gg r_0 is I=\dfrac{P}{4\pi R^2}=\dfrac{\sigma	imes4\pi r^2 T^4}{4\pi R^2}=\dfrac{\sigma r^2 T^4}{R^2}=\dfrac{r^2\sigma(t+273)^4}{R^2}.

Question 19

A black body at 1227^\circC emits radiations with maximum intensity at a wavelength of 5000 Å. If the temperature of the body is increased by 1000^\circC, the maximum intensity will be observed at

Accepted Answer

3000 Å. By Wien's law \lambda_m T= constant, so \dfrac{\lambda_m'}{\lambda_m}=\dfrac{T}{T'}. Here T=1227+273=1500 K and T'=1227+1000+273=2500 K. Hence \lambda_m'=\dfrac{1500}{2500}	imes5000=3000 Å.

Question 20

Which of the following circular rods, (given radius r and length l) each made of the same material and whose ends are maintained at the same temperature will conduct most heat?

Accepted Answer

r=2r_0; l=l_0. By conduction, H=\dfrac{\Delta Q}{\Delta t}=KA\dfrac{(T_1-T_2)}{l}, so H\propto\dfrac{r^2}{l}. (a) \dfrac{(2r_0)^2}{2l_0}=\dfrac{2r_0^2}{l_0}; (b) \dfrac{(2r_0)^2}{l_0}=\dfrac{4r_0^2}{l_0}; (c) \dfrac{r_0^2}{l_0}; (d) \dfrac{r_0^2}{2l_0}. Case (b) is the largest, so it conducts the most heat.

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