JEE Physics Thermal Properties Of Matter MCQs Set 15

Answer: (c) the temperature of the water is the same as the temperature of the remaining ice cubes

Explanation:
1. Some ice has melted, so ice and water are both present in the bowl.
2. In thermal equilibrium all parts of the system are at the same temperature, which here is the melting point 0 °C.
3. So the water and the remaining ice are at the same temperature.

Teacher's Note:
1. Whenever ice and water coexist at equilibrium, the temperature is the melting point.
2. The masses decide how much melts, not the equilibrium temperature.

Answer: (a) \( \dfrac{15a}{4} \)

Explanation:
1. For variable specific heat, \( Q=\int S\,dT \) per unit mass.
2. \( Q=\int_{1}^{2}aT^{3}\,dT=\dfrac{a}{4}\left(2^{4}-1^{4}\right) \).
3. \( Q=\dfrac{15a}{4} \).

Teacher's Note:
1. When specific heat depends on temperature, always integrate instead of multiplying.
2. Do not take an average specific heat unless it is asked for.

Answer: (b) 8200 cal

Explanation:
1. Melt the ice: \( 10\times80=800 \) cal.
2. Heat the 10 g of water from 0 °C to 100 °C: \( 10\times100=1000 \) cal.
3. Heat the calorimeter (water equivalent 10 g) from 0 °C to 100 °C: 1000 cal.
4. Vaporise the water: \( 10\times540=5400 \) cal.
5. Total \( =800+1000+1000+5400=8200 \) cal.

Teacher's Note:
1. Do not forget the calorimeter, which absorbs heat just like an equal mass of water equal to its water equivalent.
2. Add the heat for each stage: melting, heating, vaporising.

Answer: (c) 0.8 °C/s

Explanation:
1. Mass of ice \( =0.1\times100=10 \) g.
2. Rate of heat supply \( =0.1\times80=8 \) cal/s.
3. For the water, \( 8=10\times1\times\dfrac{dT}{dt} \), so \( \dfrac{dT}{dt}=0.8 \) °C/s.

Teacher's Note:
1. The same constant power first melts the ice and then heats the water.
2. Rate of temperature rise is power divided by heat capacity.

Answer: (d) 35 °C

Explanation:
1. Power \( =\dot m\,s\,\Delta T \), so \( \Delta T=\dfrac{2100}{20\times4.2}=25 \) °C.
2. Outlet temperature \( =10+25=35 \) °C.

Teacher's Note:
1. In flow problems use heat per second = mass flow rate times specific heat times temperature rise.
2. Keep the units of mass and specific heat consistent (here grams).

Answer: (b) \( 3.8\times10^{5} \) J/kg

Explanation:
1. Let the flask have water equivalent \( w \) g and the latent heat be \( L \) cal/g.
2. First stage: \( 200(30)+w(30)=50L+50(40) \), so \( 30w=50L-4000 \).
3. Second stage: \( 250(30)+w(30)=80L+80(10) \), so \( 30w=80L-6700 \).
4. Equating: \( 30L=2700 \), so \( L=90 \) cal/g.
5. \( L=90\times4.2\times10^{3}\approx3.8\times10^{5} \) J/kg.

Teacher's Note:
1. Two unknowns need two energy-balance equations.
2. Heat lost by the warm water and flask equals heat gained by the melting ice and its meltwater.

Answer: (d) 80 °C

Explanation:
1. In steady state the same heat current flows through both rods: \( \dfrac{K_1(100-\theta)}{l_1}=\dfrac{K_2(\theta-0)}{l_2} \).
2. Substituting \( K_1=2K_2 \) and \( l_2=2l_1 \): \( 2(100-\theta)=\dfrac{\theta}{2} \).
3. \( 400-4\theta=\theta \), so \( \theta=80 \) °C.

Teacher's Note:
1. Series rods carry the same heat current in steady state.
2. The junction stays closer to the temperature of the better conductor's end.

Answer: (d) 12 °C

Explanation:
1. Same heat current through both layers: \( K_A\,\Delta T_A=K_B\,\Delta T_B \) (same thickness and area).
2. \( \Delta T_A=\dfrac{K_B}{K_A}\Delta T_B=\dfrac{1}{2}\Delta T_B \).
3. \( \Delta T_A+\Delta T_B=36 \), so \( \Delta T_A=12 \) °C.

Teacher's Note:
1. The poorer conductor takes the larger share of the temperature drop.
2. Thermal resistance \( l/KA \) adds in series.

Answer: (a) \( 1.43\times10^{3} \) W

Explanation:
1. Area of each cube face \( A=(0.03)^{2}=9\times10^{-4} \) m², length \( l=0.03 \) m, \( \Delta T=80 \) K.
2. In parallel the currents add: \( H=\dfrac{(K_{Cu}+K_{Al})A\,\Delta T}{l} \).
3. \( H=\dfrac{594\times9\times10^{-4}\times80}{0.03}\approx1.43\times10^{3} \) W.

Teacher's Note:
1. Parallel conduction paths add their conductances.
2. Convert cm to metres before substituting.

Answer: (d) 0.5 min

Explanation:
1. End to end (series): length \( 2l \), area \( A \); \( H_s=\dfrac{KA\,\Delta T}{2l} \).
2. Side by side (parallel): length \( l \), area \( 2A \); \( H_p=\dfrac{2KA\,\Delta T}{l}=4H_s \).
3. The time for the same heat is one quarter: \( 2/4=0.5 \) min.

Teacher's Note:
1. Doubling the area and halving the length multiplies the heat current by 4.
2. Time for a fixed amount of heat is inversely proportional to the heat current.

Answer: (c) \( \dfrac{k_1+k_2}{2} \)

Explanation:
1. For equal areas in parallel, the total conductance is the sum of the individual conductances.
2. With equal numbers of blocks of each kind, the average conductivity is \( \dfrac{k_1+k_2}{2} \).

Teacher's Note:
1. Parallel conductors with equal areas give an arithmetic mean of conductivities.
2. Series layers of equal thickness would give the harmonic form \( \dfrac{2k_1k_2}{k_1+k_2} \).

Answer: (b) \( \dfrac{1}{4}(k_1+3k_2) \)

Explanation:
1. Heat flows along the axis, so the core and shell are in parallel.
2. Core area \( \pi R^{2} \), shell area \( \pi(4R^{2}-R^{2})=3\pi R^{2} \).
3. \( K_{eff}(4\pi R^{2})=k_1\pi R^{2}+k_2(3\pi R^{2}) \), so \( K_{eff}=\dfrac{k_1+3k_2}{4} \).

Teacher's Note:
1. Weight each conductivity by the cross-sectional area it occupies.
2. Total area is \( \pi(2R)^{2}=4\pi R^{2} \).

Answer: (c) reflected

Explanation:
1. A perfect black body absorbs all radiation falling on it and reflects none.
2. It does emit radiation characteristic of its temperature.
3. So radiation is not reflected from a black body.

Teacher's Note:
1. Reflectance of an ideal black body is zero.
2. The 'blackbody radiation' we see is emission, not reflection.

Answer: (d) the mass of the body

Explanation:
1. Radiated power \( P=e\sigma AT^{4} \) depends on the emissivity (nature of the surface), the area and the temperature.
2. It does not depend on the mass of the body.

Teacher's Note:
1. The mass matters for how fast the temperature changes, not for the power radiated.
2. Stefan's law contains area, emissivity and temperature.

Answer: (c) electromagnetic waves

Explanation:
1. Thermal radiation is mainly infrared and consists of electromagnetic waves.
2. It travels through vacuum at the speed of light, unlike sound waves.
3. Beta rays are streams of electrons, and gravity waves are not electromagnetic.

Teacher's Note:
1. Heat radiation, light and radio waves differ only in wavelength.
2. A medium is not needed for electromagnetic waves.

Answer: (c) A medium is necessary for its propagation

Explanation:
1. Thermal radiation is electromagnetic and travels at the speed of light.
2. It has particle (photon) nature as well.
3. It needs no medium, so the statement that a medium is necessary is not a property of it.

Teacher's Note:
1. The Sun's heat reaches us across empty space only because radiation needs no medium.
2. Conduction and convection need a medium.

Answer: (b) White (bright)

Explanation:
1. Emissivity is smallest for bright polished white surfaces, which reflect most radiation.
2. Black and rough surfaces emit the most.
3. So a bright white surface emits the least heat radiation.

Teacher's Note:
1. Emissivity equals absorptivity at the same temperature.
2. Shiny, light surfaces are poor emitters.

Answer: (a) \( a+r+t=1 \)

Explanation:
1. Incident radiation is partly absorbed, reflected or transmitted.
2. Hence the fractions satisfy \( a+r+t=1 \).
3. The emissive power does not enter this energy balance.

Teacher's Note:
1. Absorption, reflection and transmission fractions must add to one.
2. For an opaque body \( t=0 \) so \( a+r=1 \).

Answer: (d) 4 P

Explanation:
1. Power from the sphere \( \propto T^{4} \) and intensity at the foil \( \propto \dfrac{1}{d^{2}} \).
2. So the power received \( \propto \dfrac{T^{4}}{d^{2}} \).
3. New power \( =\dfrac{2^{4}}{2^{2}}P=4P \).

Teacher's Note:
1. Stefan's law gives the source power; the inverse square law spreads it.
2. Combine both dependences before substituting.

Answer: (c) \( \left(\dfrac{\varepsilon_P}{\varepsilon_Q}\right)^{1/4}\theta_P \)

Explanation:
1. Equal power with equal areas: \( \varepsilon_P\sigma A\theta_P^{4}=\varepsilon_Q\sigma A\theta_Q^{4} \).
2. \( \theta_Q^{4}=\dfrac{\varepsilon_P}{\varepsilon_Q}\theta_P^{4} \).
3. \( \theta_Q=\left(\dfrac{\varepsilon_P}{\varepsilon_Q}\right)^{1/4}\theta_P \).

Teacher's Note:
1. A body with lower emissivity must be hotter to radiate the same power.
2. Take the fourth root at the end.

Answer: (c) \( \dfrac{E}{16} \)

Explanation:
1. Temperatures in kelvin are 546 K and 273 K.
2. \( \dfrac{E'}{E}=\left(\dfrac{273}{546}\right)^{4}=\dfrac{1}{16} \).
3. So the rate at 0 °C is \( E/16 \).

Teacher's Note:
1. Always convert to kelvin before using Stefan's law.
2. Halving the absolute temperature reduces the power by 16.

Answer: (a) \( \dfrac{256}{81} \)

Explanation:
1. Wien's law \( \lambda_mT=\text{const} \), so \( T'=\dfrac{4}{3}T \).
2. Stefan's law \( P\propto T^{4} \): \( \dfrac{P'}{P}=\left(\dfrac{4}{3}\right)^{4}=\dfrac{256}{81} \).

Teacher's Note:
1. Wien's law gives the new temperature; Stefan's law gives the new power.
2. A shorter peak wavelength means a hotter body.

Answer: (a) Both will emit equal amounts of radiation per unit time

Explanation:
1. Radiated power depends on surface area, emissivity and temperature, which are identical for the two spheres.
2. So both emit the same power.
3. Their masses differ, so the rates of fall of temperature differ.

Teacher's Note:
1. Radiated power depends on the surface, not on the volume or mass.
2. Rate of cooling also depends on the heat capacity.

Answer: (d) all of these

Explanation:
1. Rate of fall of temperature \( \dfrac{dT}{dt}=\dfrac{e\sigma A(T^{4}-T_0^{4})}{ms} \).
2. It depends on area \( A \), mass \( m \) and specific heat \( s \) (and on emissivity and temperatures).

Teacher's Note:
1. Cooling rate equals radiated power divided by heat capacity.
2. Large area speeds up cooling; large mass or specific heat slows it down.