JEE Physics Thermal Properties Of Matter MCQs Set 08

Answer: (b) 375 g

Explanation:
1. Heat lost by the hot liquid \( =m(0.2)(40-32)=1.6m \).
2. Heat gained by the cold liquid \( =100(0.5)(32-20)=600 \).
3. \( 1.6m=600\Rightarrow m=375\ \text{g} \).

Teacher's Note:
1. Heat lost equals heat gained in a mixture.
2. Use the specific heats given in cal/g°C.

Answer: (a) \( 0^\circ\text{C} \)

Explanation:
1. Heat released by water cooling to 0°C: \( 300\times1\times25=7500\ \text{cal} \).
2. Heat needed to melt all the ice: \( 100\times80=8000\ \text{cal} \).
3. The water cannot melt all the ice, so some ice remains and the mixture stays at 0°C.

Teacher's Note:
1. Compare the available heat with the heat needed to melt all the ice.
2. A mixture of ice and water coexists at 0°C.

Answer: (c) equal for both balls

Explanation:
1. Heat produced equals the kinetic energy: \( \tfrac12mv^2=ms\,\Delta T \).
2. \( \Delta T=\frac{v^2}{2s} \), independent of mass.
3. So the temperature rise is the same for both.

Teacher's Note:
1. Mass cancels from both sides of the energy equation.
2. The same speed and the same metal means the same temperature rise.

Answer: (c) (H2 − H1) represents the latent heat of fusion of the substance

Explanation:
1. Horizontal portions are changes of state at constant temperature: AB is melting at T1 and CD is boiling at T2.
2. The heat supplied during melting of 1 kg is \( H_2-H_1 \), the latent heat of fusion.
3. T1 is the melting point, BC is the heating of the liquid, and the latent heat of vaporisation is \( H_4-H_3 \). Only (c) is correct.

Teacher's Note:
1. A flat part of a heating curve means a change of state.
2. Latent heat is the length of that flat part along the heat axis.

Answer: (d) C

Explanation:
1. With the same heat supply rate P and the same mass, \( \frac{dT}{dt}=\frac{P}{mc} \).
2. The slope is inversely proportional to the specific heat.
3. C has the smallest slope, so it has the largest specific heat.

Teacher's Note:
1. A smaller temperature rise for the same heat means a larger specific heat.
2. Slope of the T-t graph is P/(mc).

Answer: (c) specific heat of liquids

Explanation:
1. In the cooling method, equal volumes of a liquid and of water (or two liquids) are allowed to cool under the same conditions and the times of cooling through the same range are compared.
2. Newton's law of cooling relates the rate of fall of temperature to \( \frac{ms\,\Delta T}{\Delta t} \), which gives the ratio of specific heats of the liquids.
3. So it is used to compare the specific heats of liquids.

Teacher's Note:
1. The method compares cooling times of two liquids.
2. It is not used for latent heats.

Answer: (b) \( T_1\lt T_2\lt T_3 \)

Explanation:
1. By Newton's law the rate of cooling is proportional to the excess temperature over the surroundings.
2. As the water gets cooler, its excess temperature decreases and it loses heat more slowly.
3. So each successive 5°C fall takes longer: \( T_1\lt T_2\lt T_3 \).

Teacher's Note:
1. The rate of cooling slows down as the body nears the room temperature.
2. The same temperature drop therefore takes longer each time.

Answer: (a) \( \frac{81}{64}E \)

Explanation:
1. Rate of emission \( \propto AT^4 \). Area becomes \( \frac14 \) of the original.
2. Temperatures: \( 127^\circ\text{C}=400\ \text{K} \) and \( 327^\circ\text{C}=600\ \text{K} \).
3. \( E'=E\times\frac14\times\left(\frac{600}{400}\right)^4=E\times\frac14\times\frac{81}{16}=\frac{81}{64}E \).

Teacher's Note:
1. Convert to kelvin before using the Stefan-Boltzmann law.
2. Halving both length and breadth reduces the area to a quarter.

Answer: (d) \( \frac1{81} \)

Explanation:
1. Temperatures in kelvin: \( -73^\circ\text{C}=200\ \text{K} \) and \( 327^\circ\text{C}=600\ \text{K} \).
2. Emissive power \( \propto T^4 \): \( \frac{E_1}{E_2}=\left(\frac{200}{600}\right)^4=\frac1{81} \).

Teacher's Note:
1. Always use absolute temperature in radiation problems.
2. Emissive power varies as the fourth power of the temperature.

Answer: (c) 30 : 1

Explanation:
1. Wien's law: \( \lambda_mT=\text{constant} \).
2. \( \frac{T_{sun}}{T_{moon}}=\frac{\lambda_{moon}}{\lambda_{sun}}=\frac{4200}{140}=30 \).

Teacher's Note:
1. The hotter body has the shorter peak wavelength.
2. The ratio of temperatures is the inverse of the ratio of wavelengths.

Answer: (b) 20 cal/s

Explanation:
1. By Newton's law, the rate of loss of heat is proportional to the excess temperature over the surroundings.
2. Excess temperatures: \( 80-20=60 \) and \( 40-20=20 \).
3. \( \text{rate}=60\times\frac{20}{60}=20\ \text{cal/s} \).

Teacher's Note:
1. Use the temperature difference, not the absolute temperature.
2. Rate is directly proportional to the excess temperature.

Answer: (a) 0.33 μm

Explanation:
1. Wien's displacement law: \( \lambda_mT=\text{constant} \).
2. \( \lambda_2=4\times\frac{200}{2400}=0.33\ \mu\text{m} \).

Teacher's Note:
1. The peak wavelength is inversely proportional to the absolute temperature.
2. A twelve times hotter body peaks at one-twelfth of the wavelength.

Answer: (d) 0.64

Explanation:
1. Rate of heat flow \( =\frac{KA\,\Delta T}{L}\propto\frac1L \) (same K, A and ΔT).
2. If the diameter is d, the straight rod has length d and the semicircular rod has length \( \frac{\pi d}{2} \).
3. \( \frac{Q_{semi}}{Q_{straight}}=\frac{d}{\pi d/2}=\frac2\pi\approx0.64 \).

Teacher's Note:
1. Longer rods conduct heat more slowly under the same temperature difference.
2. The length of a semicircle is π times the radius.

Answer: (b) \( \frac{A_1}{A_2}=\frac{K_2}{K_1} \)

Explanation:
1. Rate of heat flow \( =\frac{KA\,\Delta T}{L} \) is the same for both, with the same L and ΔT.
2. \( K_1A_1=K_2A_2\Rightarrow\frac{A_1}{A_2}=\frac{K_2}{K_1} \).

Teacher's Note:
1. The product KA must be equal.
2. A poorer conductor needs a larger area.

Answer: (c) \( K=\infty \)

Explanation:
1. In the steady state, a temperature difference along the rod is \( \Delta T=\frac{Q\,L}{KA} \) for a given heat flow.
2. The difference vanishes only when the thermal conductivity is infinitely large.

Teacher's Note:
1. Larger conductivity means a smaller temperature difference for the same heat current.
2. Uniform temperature is an ideal limit.

Answer: (b) \( \frac{K_1+3K_2}{4} \)

Explanation:
1. The two materials conduct heat in parallel along the length.
2. Areas: core \( \pi R^2 \), shell \( \pi(4R^2-R^2)=3\pi R^2 \), total \( 4\pi R^2 \).
3. \( K_{eff}\cdot4\pi R^2=K_1\pi R^2+K_2\cdot3\pi R^2\Rightarrow K_{eff}=\frac{K_1+3K_2}{4} \).

Teacher's Note:
1. Parallel paths add their conductances (K × A).
2. Divide by the total area to get the effective K.

Answer: (d) \( K_A=\frac{K_B}{4} \)

Explanation:
1. Equal rates: \( K_AA_A=K_BA_B \) (same L and ΔT).
2. Areas: \( A_A=4A_B \), so \( K_A\times4A_B=K_BA_B\Rightarrow K_A=\frac{K_B}{4} \).

Teacher's Note:
1. Area is proportional to the square of the radius.
2. The wire with a larger area needs a smaller conductivity for the same heat flow.

Answer: (a) first decreases and then increases

Explanation:
1. Water shows anomalous expansion: its volume decreases as it is heated from 0°C to 4°C (its density is maximum at 4°C).
2. Above 4°C it expands normally, so the volume increases up to 10°C.

Teacher's Note:
1. Water has its maximum density at 4°C.
2. Ice floats because of this anomaly.