JEE Physics Thermal Properties Of Matter MCQs Set 13

Answer: (b) 64 J

Explanation:
1. Area \( A=0.1\times0.1=0.01\text{ m}^2 \), thickness \( l=0.01\text{ m} \).
2. \( \dfrac{Q}{t}=\dfrac{kA\Delta T}{l}=\dfrac{0.80\times0.01\times80}{0.01}=64\text{ W} \), so 64 J flows each second.

Teacher's Note:
1. The heat flows across the thickness, so the area is that of the larger face.
2. Convert centimetres to metres first.

Answer: (d) \( \dfrac{t_1+t_2+t_3}{\frac{t_1}{k_1}+\frac{t_2}{k_2}+\frac{t_3}{k_3}} \)

Explanation:
1. Slabs in series carry the same heat current, so thermal resistances add: \( \dfrac{t_1+t_2+t_3}{kA}=\dfrac{t_1}{k_1A}+\dfrac{t_2}{k_2A}+\dfrac{t_3}{k_3A} \).
2. Solving gives the effective conductivity above.

Teacher's Note:
1. Resistance of a slab is thickness divided by (conductivity times area).
2. Only for equal thicknesses does this reduce to the harmonic mean.

Answer: (c) \( \dfrac{15}{\pi}\approx4.8\text{ W/m}^\circ\text{C} \)

Explanation:
1. Radial conduction through a spherical shell: \( H=\dfrac{4\pi k\,r_1r_2(T_1-T_2)}{r_2-r_1} \).
2. \( k=\dfrac{160\times0.15}{4\pi\times0.05\times0.20\times40}=\dfrac{24}{1.6\pi}=\dfrac{15}{\pi}\approx4.8 \).

Teacher's Note:
1. The area changes with radius, so the simple slab formula cannot be used.
2. Add the resistances of thin shells to derive the formula.

Answer: (a) 12°C/s

Explanation:
1. Heat entering at A minus heat leaving at B: \( kA(g_A-g_B)=200\times10^{-4}\times(500-260)=4.8\text{ W} \) (gradients \( 500\text{ and }260^\circ\text{C/m} \)).
2. Rate of rise \( =\dfrac{4.8}{0.40}=12^\circ\text{C/s} \).

Teacher's Note:
1. Convert the gradients to °C per metre.
2. The difference between the two heat currents warms the section.

Answer: (d) \( \dfrac{q_1-q_2}{q_1} \)

Explanation:
1. Energy absorbed \( =q_1-q_2 \), so absorptive power \( a=\dfrac{q_1-q_2}{q_1} \).
2. By Kirchhoff's law emissivity equals absorptive power at the same temperature.

Teacher's Note:
1. Absorbed + reflected + transmitted = incident.
2. A good absorber is a good emitter.

Answer: (b) 6000 K and 4000 K

Explanation:
1. \( T=\dfrac{b}{\lambda_m} \): blue \( \dfrac{3\times10^{-3}}{5\times10^{-7}}=6000\text{ K} \); red \( \dfrac{3\times10^{-3}}{7.5\times10^{-7}}=4000\text{ K} \).

Teacher's Note:
1. Shorter wavelength means a hotter body.
2. Convert 0.3 cm K to m K.

Answer: (a) \( \dfrac1{90} \)

Explanation:
1. Heat is lost at a constant power \( P \). Cooling: \( mc\times3=P\cdot1\text{ min} \). Solidifying: \( mL=P\times30\text{ min} \).
2. Dividing: \( \dfrac cL=\dfrac{1}{3\times30}=\dfrac1{90}\text{ K}^{-1} \).

Teacher's Note:
1. The surroundings take heat at the same rate throughout.
2. Express both heats per minute.

Answer: (c) 7 minutes

Explanation:
1. Excess temperatures: 40, 30, 20. \( \ln\dfrac{40}{30}=k\times5\Rightarrow k=0.0575\text{ min}^{-1} \).
2. Next interval: \( t=\dfrac1k\ln\dfrac{30}{20}=\dfrac{0.405}{0.0575}\approx7.0\text{ min} \).

Teacher's Note:
1. Newton's law gives an exponential decay of the temperature excess over the surroundings.
2. The same fall takes longer as the body gets closer to the surrounding temperature.

Answer: (b) 2.9 : 1

Explanation:
1. Heat loss \( \propto r^2 \); heat capacity \( =\tfrac43\pi r^3\rho s \), so rate of fall \( \propto\dfrac{1}{r\rho s} \).
2. \( \dfrac{(dT/dt)_{Al}}{(dT/dt)_{Cu}}=\dfrac{r_{Cu}\rho_{Cu}s_{Cu}}{r_{Al}\rho_{Al}s_{Al}}=\dfrac{2\times3.4\times390}{900}=2.9 \).

Teacher's Note:
1. The ratio of the heat loss rates alone is 1 : 4, but the heat capacity of copper is much greater.
2. A smaller sphere cools faster.

Answer: (d) 180 W and 9 W

Explanation:
1. At the start the block is at room temperature, so no heat is lost: \( P=C\dfrac{dT}{dt}=90\times2=180\text{ W} \).
2. Just after switching off (35°C, excess 10°C) the radiated power is \( 90\times0.2=18\text{ W} \); it is proportional to the excess temperature, so at 30°C (excess 5°C) it is \( 9\text{ W} \).

Teacher's Note:
1. Radiated power is proportional to the excess over room temperature.
2. The heater power equals the sum of heat stored and heat lost.

Answer: (c) \( \dfrac{100}{19}\text{ s} \)

Explanation:
1. Heat stored \( =90\times10=900\text{ J} \).
2. Net power \( =180-9=171\text{ W} \), so \( t=\dfrac{900}{171}=\dfrac{100}{19}\text{ s}\approx5.3\text{ s} \).

Teacher's Note:
1. The heater has to supply both the stored energy and the average loss.
2. Without any loss it would take 5 s.

Answer: (a) \( \dfrac{mL\ln(r_2/r_1)}{100\pi k\ell} \)

Explanation:
1. Radial heat flow \( H=\dfrac{2\pi k\ell(50-0)}{\ln(r_2/r_1)}=\dfrac{100\pi k\ell}{\ln(r_2/r_1)} \).
2. Time to supply \( mL \): \( t=\dfrac{mL}{H}=\dfrac{mL\ln(r_2/r_1)}{100\pi k\ell} \).

Teacher's Note:
1. The ice stays at 0°C until it has all melted.
2. Heat enters only through the curved wall.

Answer: (d) 887 J

Explanation:
1. \( E=\sigma AT^4=6\times10^{-8}\times1.6\times(310)^4=6\times10^{-8}\times1.6\times9.235\times10^{9}\approx887\text{ W} \), i.e. 887 J each second.

Teacher's Note:
1. Temperature must be in kelvin: 37 + 273 = 310 K.
2. This is the emitted energy; the body also absorbs from surroundings.

Answer: (b) 92 W

Explanation:
1. \( P=e\sigma A(T^4-T_0^4)=0.97\times5.67\times10^{-8}\times2\times(301^4-293^4) \).
2. \( =1.1\times10^{-7}\times8.38\times10^{8}\approx92\text{ W} \).

Teacher's Note:
1. Use kelvin temperatures and subtract the fourth powers.
2. The small temperature difference makes the net loss modest.

Answer: (a) 9 kW

Explanation:
1. Resistance per unit area \( \sum\dfrac{t}{k}=\dfrac{0.025}{0.125}+\dfrac{0.01}{1.5}+\dfrac{0.25}{1.0}=0.4567\text{ m}^2\text{K/W} \).
2. Power \( =\dfrac{A\Delta T}{\sum t/k}=\dfrac{137\times30}{0.4567}=9.0\times10^3\text{ W} \).

Teacher's Note:
1. Thermal resistances of layers in series add.
2. In steady state the heater replaces the heat lost through the walls.

Answer: (c) \( (\Delta T)_0e^{-\lambda t},\ \lambda=\left(\dfrac1{C_1}+\dfrac1{C_2}\right)\dfrac{SK}{\ell} \)

Explanation:
1. Heat current \( H=\dfrac{KS}{\ell}\Delta T \) goes from the hotter to the cooler chunk: \( C_1\dfrac{dT_1}{dt}=-H,\ C_2\dfrac{dT_2}{dt}=+H \).
2. So \( \dfrac{d(\Delta T)}{dt}=-\left(\dfrac1{C_1}+\dfrac1{C_2}\right)\dfrac{KS}{\ell}\Delta T \), giving the exponential decay above.

Teacher's Note:
1. The temperature difference decays exponentially, just as in Newton's law of cooling.
2. The smaller heat capacity dominates the decay rate.

Answer: (b) 12°C

Explanation:
1. Same heat current in both layers: \( \dfrac{k_A\Delta T_A}{l}=\dfrac{k_B\Delta T_B}{l} \), so \( \Delta T_A=\dfrac{\Delta T_B}{2} \).
2. With \( \Delta T_A+\Delta T_B=36 \), \( \Delta T_A=12^\circ\text{C} \).

Teacher's Note:
1. The better conductor has the smaller temperature drop.
2. Series layers carry equal current.

Answer: (d) −30°C

Explanation:
1. For ice thickness \( x \): \( \rho L\,\dfrac{dx}{dt}=\dfrac{K\theta}{x} \) where \( \theta \) is the temperature drop.
2. Integrating: \( \dfrac{x_2^2-x_1^2}{2}=\dfrac{K\theta t}{\rho L}\Rightarrow6=\dfrac{4\times10^{-3}\times\theta\times3600}{0.9\times80} \).
3. So \( \theta=30^\circ\text{C} \) and the air is at \( -30^\circ\text{C} \).

Teacher's Note:
1. The freezing rate falls as the ice gets thicker.
2. Convert the time to seconds.

Answer: (c) \( \dfrac{2R_1R_2}{R_1+R_2} \)

Explanation:
1. For steady radial flow, \( \theta(r)-\theta_1\propto\dfrac1{R_1}-\dfrac1r \).
2. Half-way: \( \dfrac1{R_1}-\dfrac1r=\dfrac12\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\Rightarrow\dfrac1r=\dfrac12\left(\dfrac1{R_1}+\dfrac1{R_2}\right) \).
3. Hence \( r=\dfrac{2R_1R_2}{R_1+R_2} \).

Teacher's Note:
1. The temperature falls as 1/r in radial conduction through a sphere.
2. It is the harmonic mean of the two radii.

Answer: (a) \( P\propto r^2\text{ and }R\propto\dfrac1r \)

Explanation:
1. Emitted power \( P=\sigma T^4\cdot4\pi r^2\propto r^2 \).
2. Heat capacity \( \propto r^3 \), so rate of cooling \( R=\dfrac{P}{ms}\propto\dfrac{r^2}{r^3}=\dfrac1r \).

Teacher's Note:
1. Small bodies cool faster than large bodies at the same temperature.
2. Power goes with area, heat capacity with volume.

Answer: (d) 86°C and 57°C

Explanation:
1. Resistances are proportional to \( \dfrac1K \): copper \( 1 \), aluminium \( 2 \), brass \( 4 \) (in units of \( l/(K_{Cu}A) \)); total 7.
2. Drop across copper \( =\dfrac17\times100=14.3^\circ\text{C} \), across aluminium \( \dfrac27\times100=28.6^\circ\text{C} \).
3. Junctions: \( 100-14.3=85.7^\circ\text{C} \) and \( 85.7-28.6=57.1^\circ\text{C} \).

Teacher's Note:
1. The same heat current flows through all three rods.
2. The larger the conductivity the smaller the temperature drop.