JEE Physics Thermal Properties Of Matter MCQs Set 21

Answer: (b) 20 g

Explanation:
1. Let \( m \) gram evaporate. The heat needed is taken from the water that freezes: \( m(2.10\times10^6)=(150-m)(3.36\times10^5) \).
2. \( m(2.436\times10^6)=150(3.36\times10^5) \Rightarrow m\approx20.7 g \), closest to \( 20 g \).

Teacher's Note:
1. In calorimetry write heat lost = heat gained, and check first whether the ice melts completely or only partly before assuming a final temperature.
2. Keep the units consistent (cal with \( g, J \) with kg) and add the latent heat only for the mass that actually changes state.

Answer: (c) \( 80^\circ\text{C} \)

Explanation:
1. First case: \( 100c_A(10)=50c_B(15)\Rightarrow c_B=\dfrac43c_A \).
2. Second case: \( 100c_A(100-T)=50\cdot\dfrac43c_A(T-50)\Rightarrow 300(100-T)=200(T-50)\Rightarrow T=80^\circ C \).

Teacher's Note:
1. In calorimetry write heat lost = heat gained, and check first whether the ice melts completely or only partly before assuming a final temperature.
2. Keep the units consistent (cal with \( g, J \) with kg) and add the latent heat only for the mass that actually changes state.

Answer: (a) 20%

Explanation:
1. \( 0.1(400)(500-T)=(800+0.5\times4200)(T-30)\Rightarrow 40(500-T)=2900(T-30) \).
2. \( T\approx36.4^\circ C, a \) rise of 6.4 \( ^\circ C \) on 30 \( ^\circ C \), i.e. about 21%, closest to 20%.

Teacher's Note:
1. In calorimetry write heat lost = heat gained, and check first whether the ice melts completely or only partly before assuming a final temperature.
2. Keep the units consistent (cal with \( g, J \) with kg) and add the latent heat only for the mass that actually changes state.

Answer: (b) 40 g

Explanation:
1. Heat lost by water \( =50(4.2)(40)=8400 J \).
2. Let \( m \) be the ice added: \( m(2.1)(20)+(m-20)(334)=8400\Rightarrow376m=15080\Rightarrow m\approx40 g \).

Teacher's Note:
1. In calorimetry write heat lost = heat gained, and check first whether the ice melts completely or only partly before assuming a final temperature.
2. Keep the units consistent (cal with \( g, J \) with kg) and add the latent heat only for the mass that actually changes state.

Answer: (c) \( 916\ \text{J kg}^{-1}\text{K}^{-1} \)

Explanation:
1. Rise in temperature of water and calorimeter: \( 21.5-8.4=13.1 K \). Heat gained \( =0.24(4180)(13.1)+0.128(394)(13.1)\approx13803 J \).
2. Heat lost by the metal \( =0.192\,c\,(100-21.5)=15.07c \), so \( c\approx916 J kg ^{-1} K ^{-1} \).

Teacher's Note:
1. In calorimetry write heat lost = heat gained, and check first whether the ice melts completely or only partly before assuming a final temperature.
2. Keep the units consistent (cal with \( g, J \) with kg) and add the latent heat only for the mass that actually changes state.

Answer: (a) \( 885^\circ\text{C} \)

Explanation:
1. Heat lost by ball \( =100(0.1)(T-75) \).
2. Heat gained \( =100(0.1)(45)+170(1)(45)=8100 \) cal, so \( T-75=810\Rightarrow T=885^\circ C \).

Teacher's Note:
1. In calorimetry write heat lost = heat gained, and check first whether the ice melts completely or only partly before assuming a final temperature.
2. Keep the units consistent (cal with \( g, J \) with kg) and add the latent heat only for the mass that actually changes state.

Answer: (b) 2067

Explanation:
1. Heat produced by the device in \( 3 h =3000\times10800=3.24\times10^7 J \).
2. The water can absorb at most \( 120(4200)(30-10)=1.008\times10^7 J \); the cooler must remove the rest: \( P\times10800=2.232\times10^7\Rightarrow P\approx2067 W \).

Teacher's Note:
1. In calorimetry write heat lost = heat gained, and check first whether the ice melts completely or only partly before assuming a final temperature.
2. Keep the units consistent (cal with \( g, J \) with kg) and add the latent heat only for the mass that actually changes state.

Answer: (d) 8 min 20 s

Explanation:
1. Heat needed \( =2(4200)(50)=4.2\times10^5 J \).
2. Net power \( =1000-160=840 W \), so \( t=\dfrac{4.2\times10^5}{840}=500 s = 8 \min 20 s \).

Teacher's Note:
1. In calorimetry write heat lost = heat gained, and check first whether the ice melts completely or only partly before assuming a final temperature.
2. Keep the units consistent (cal with \( g, J \) with kg) and add the latent heat only for the mass that actually changes state.

Answer: (a) Five parts: a steep rise, a short flat part at 0 \(^\circ\)C, a less steep rise, a much longer flat part at 100 \(^\circ\)C, and a final steep rise

Explanation:
1. Slope of the \( T-Q \) graph is \( 1/(mc) \): ice and steam have small \( c \) so steep lines; water has the largest \( c \) so the least steep line.
2. \( L_v\approx540 \) cal/g is much more than \( L_f=80 \) cal/g, so the flat part at 100 \( ^\circ C \) is much longer than the one at 0 \( ^\circ C \).

Teacher's Note:
1. In calorimetry write heat lost = heat gained, and check first whether the ice melts completely or only partly before assuming a final temperature.
2. Keep the units consistent (cal with \( g, J \) with kg) and add the latent heat only for the mass that actually changes state.

Answer: (b) 0.130

Explanation:
1. Heat gained \( =(1100+20)(1)(65)=72800 \) cal.
2. Heat lost by steam \( =m(540)+m(1)(100-80)=560m \), so \( m=130 g = 0.130 \) kg.

Teacher's Note:
1. In calorimetry write heat lost = heat gained, and check first whether the ice melts completely or only partly before assuming a final temperature.
2. Keep the units consistent (cal with \( g, J \) with kg) and add the latent heat only for the mass that actually changes state.

Answer: (d) 8

Explanation:
1. Heat to bring \( m g \) of ice to 0 \( ^\circ C =m(2.1)(5)=10.5m J \) (specific heat \( 2.1 J g ^{-1} K ^{-1} \)).
2. Melting \( 1 g \) needs \( 10^{-3}\times3.36\times10^5=336 J \). So \( 10.5m+336=420\Rightarrow m=8 g \).

Teacher's Note:
1. In calorimetry write heat lost = heat gained, and check first whether the ice melts completely or only partly before assuming a final temperature.
2. Keep the units consistent (cal with \( g, J \) with kg) and add the latent heat only for the mass that actually changes state.

Answer: (a) 5.5 minutes

Explanation:
1. Power received \( =1400\times0.2=280 W \); heat needed \( =0.28\times3.3\times10^5=92400 J \).
2. \( t=\dfrac{92400}{280}=330 s = 5.5 \) minutes.

Teacher's Note:
1. In calorimetry write heat lost = heat gained, and check first whether the ice melts completely or only partly before assuming a final temperature.
2. Keep the units consistent (cal with \( g, J \) with kg) and add the latent heat only for the mass that actually changes state.

Answer: (b) \( \dfrac{Pt}{M} \)

Explanation:
1. To stay molten the power input just balances the heat loss, so the loss rate is \( P \) watts.
2. On solidifying, the heat lost in \( t \) seconds is \( Pt=ML \), so \( L=\dfrac{Pt}{M} \).

Teacher's Note:
1. In calorimetry write heat lost = heat gained, and check first whether the ice melts completely or only partly before assuming a final temperature.
2. Keep the units consistent (cal with \( g, J \) with kg) and add the latent heat only for the mass that actually changes state.

Answer: (d) a mixture of solid and liquid (melting in progress)

Explanation:
1. A horizontal part on a heating curve means the heat supplied is used as latent heat while the temperature stays constant.
2. The first horizontal part after the solid state is melting, so at \( P \) solid and liquid coexist.

Teacher's Note:
1. In calorimetry write heat lost = heat gained, and check first whether the ice melts completely or only partly before assuming a final temperature.
2. Keep the units consistent (cal with \( g, J \) with kg) and add the latent heat only for the mass that actually changes state.

Answer: (a) 273 K

Explanation:
1. Heat released when the steam condenses and cools to \( 273 K =50(540)+50(1)(100)=32000 \) cal. Heat to warm ice to \( 273 K =450(0.5)(20)=4500 \) cal.
2. Remaining 27500 cal is less than the 36000 cal needed to melt all the ice, so the mixture stays at \( 273 K \).

Teacher's Note:
1. In calorimetry write heat lost = heat gained, and check first whether the ice melts completely or only partly before assuming a final temperature.
2. Keep the units consistent (cal with \( g, J \) with kg) and add the latent heat only for the mass that actually changes state.

Answer: (c) 12 g

Explanation:
1. Heat gained by water \( =100(1)(66)=6600 \) cal.
2. Steam gives \( m(540)+m(1)(100-90)=550m \), so \( m=12 g \).

Teacher's Note:
1. In calorimetry write heat lost = heat gained, and check first whether the ice melts completely or only partly before assuming a final temperature.
2. Keep the units consistent (cal with \( g, J \) with kg) and add the latent heat only for the mass that actually changes state.

Answer: (d) 410 m/s

Explanation:
1. 75% of the kinetic energy heats and melts the bullet: \( 0.75\times\tfrac12v^2=[0.03(300)+6]\times4.2\times10^3 \) J/kg \( =6.3\times10^4 \).
2. \( v^2=1.68\times10^5\Rightarrow v\approx410 m/s \).

Teacher's Note:
1. In calorimetry write heat lost = heat gained, and check first whether the ice melts completely or only partly before assuming a final temperature.
2. Keep the units consistent (cal with \( g, J \) with kg) and add the latent heat only for the mass that actually changes state.

Answer: (a) \( 3F/(\pi r^2YT) \)

Explanation:
1. Length unchanged: thermal strain = compressive strain, \( \alpha T=\dfrac{F}{\pi r^2Y}\Rightarrow\alpha=\dfrac{F}{\pi r^2YT} \).
2. \( \gamma=3\alpha=\dfrac{3F}{\pi r^2YT} \).

Teacher's Note:
1. For thermal expansion use the linear coefficient \( \alpha \) for length, \( 2 \alpha \) for area and \( 3 \alpha \) for volume, valid for small temperature changes.
2. If expansion is prevented, the thermal strain \( \alpha \) times the temperature change equals the mechanical strain, so stress is \( Y \alpha \) times the temperature change.

Answer: (c) \( 230^\circ\text{C} \)

Explanation:
1. Same initial length and same final length: \( \alpha_A(150)=\alpha_B(T-30) \).
2. \( T-30=150\times\dfrac43=200\Rightarrow T=230^\circ C \).

Teacher's Note:
1. For thermal expansion use the linear coefficient \( \alpha \) for length, \( 2 \alpha \) for area and \( 3 \alpha \) for volume, valid for small temperature changes.
2. If expansion is prevented, the thermal strain \( \alpha \) times the temperature change equals the mechanical strain, so stress is \( Y \alpha \) times the temperature change.

Answer: (d) \( \dfrac{p}{3\alpha K} \)

Explanation:
1. Volume strain due to pressure \( =\dfrac pK \).
2. Thermal volume strain \( =3\alpha\Delta T \). Setting them equal, \( \Delta T=\dfrac{p}{3\alpha K} \).

Teacher's Note:
1. For thermal expansion use the linear coefficient \( \alpha \) for length, \( 2 \alpha \) for area and \( 3 \alpha \) for volume, valid for small temperature changes.
2. If expansion is prevented, the thermal strain \( \alpha \) times the temperature change equals the mechanical strain, so stress is \( Y \alpha \) times the temperature change.

Answer: (a) 25 \(^\circ\)C, \( \alpha=1.85\times10^{-5}/^\circ \)C

Explanation:
1. Time lost per day \( =\tfrac12\alpha(T-T_0)\times86400 \). So \( \dfrac{12}{-4}=\dfrac{40-T_0}{20-T_0}\Rightarrow T_0=25^\circ C \).
2. At 40 \( ^\circ C \): \( 12=\tfrac12\alpha(15)(86400)\Rightarrow\alpha=1.85\times10^{-5}/^\circ C \).

Teacher's Note:
1. For thermal expansion use the linear coefficient \( \alpha \) for length, \( 2 \alpha \) for area and \( 3 \alpha \) for volume, valid for small temperature changes.
2. If expansion is prevented, the thermal strain \( \alpha \) times the temperature change equals the mechanical strain, so stress is \( Y \alpha \) times the temperature change.

Answer: (c) \( \dfrac{\alpha_s}{\alpha_a+\alpha_s} \)

Explanation:
1. Equal increase: \( l_1\alpha_at=l_2\alpha_st\Rightarrow\dfrac{l_1}{l_2}=\dfrac{\alpha_s}{\alpha_a} \).
2. \( \dfrac{l_1}{l_1+l_2}=\dfrac{\alpha_s}{\alpha_s+\alpha_a} \).

Teacher's Note:
1. For thermal expansion use the linear coefficient \( \alpha \) for length, \( 2 \alpha \) for area and \( 3 \alpha \) for volume, valid for small temperature changes.
2. If expansion is prevented, the thermal strain \( \alpha \) times the temperature change equals the mechanical strain, so stress is \( Y \alpha \) times the temperature change.

Answer: (d) (b) and (d) only

Explanation:
1. For a thin bimetallic strip of thickness \( d \) the radius is \( R=\dfrac{d}{(\alpha_B-\alpha_C)\Delta T} \) (up to a numerical factor).
2. So \( R \) is inversely proportional to both \( \Delta T \) and \( |\alpha_B-\alpha_C| \): \( (b) \) and \( (d) \).

Teacher's Note:
1. For thermal expansion use the linear coefficient \( \alpha \) for length, \( 2 \alpha \) for area and \( 3 \alpha \) for volume, valid for small temperature changes.
2. If expansion is prevented, the thermal strain \( \alpha \) times the temperature change equals the mechanical strain, so stress is \( Y \alpha \) times the temperature change.

Answer: (b) \( \gamma_l=2\alpha_s \)

Explanation:
1. For floating, depth \( h=a\dfrac{\rho_s}{\rho_l} \). After heating \( a'=a(1+\alpha_s\Delta T) \), \( \rho_s'=\dfrac{\rho_s}{1+3\alpha_s\Delta T} \), \( \rho_l'=\dfrac{\rho_l}{1+\gamma_l\Delta T} \).
2. \( h'=h \) requires \( (1+\alpha_s\Delta T)(1+\gamma_l\Delta T)=1+3\alpha_s\Delta T \), so \( \gamma_l\approx2\alpha_s \).

Teacher's Note:
1. For thermal expansion use the linear coefficient \( \alpha \) for length, \( 2 \alpha \) for area and \( 3 \alpha \) for volume, valid for small temperature changes.
2. If expansion is prevented, the thermal strain \( \alpha \) times the temperature change equals the mechanical strain, so stress is \( Y \alpha \) times the temperature change.

Answer: (c) \( 1.1\times10^{11}\ \text{N/m}^2 \)

Explanation:
1. Copper expands by \( 1.7\times10^{-5}(0.3)(100)=0.51 \) mm, so the other rod \( (0.7 m) \) expands by \( 1.40 \) mm: \( \alpha_2=\dfrac{1.40\times10^{-3}}{0.7\times100}=2.0\times10^{-5} / ^\circ C \).
2. If neither rod may change length, both carry the same stress: \( Y_1\alpha_1=Y_2\alpha_2\Rightarrow Y_2=\dfrac{1.3\times10^{11}\times1.7}{2.0}\approx1.1\times10^{11} N/m ^2 \).

Teacher's Note:
1. For thermal expansion use the linear coefficient \( \alpha \) for length, \( 2 \alpha \) for area and \( 3 \alpha \) for volume, valid for small temperature changes.
2. If expansion is prevented, the thermal strain \( \alpha \) times the temperature change equals the mechanical strain, so stress is \( Y \alpha \) times the temperature change.

Answer: (d) \( \dfrac{(w_2-w_1)+\beta(w_0-w_1)(T_2-T_1)}{(w_0-w_2)(T_2-T_1)} \)

Explanation:
1. Buoyant force at \( T_1 \): \( w_0-w_1=V\rho g \). At \( T_2 \): \( w_0-w_2=V(1+\beta\Delta T)\dfrac{\rho g}{1+\gamma\Delta T} \).
2. Dividing: \( \dfrac{w_0-w_2}{w_0-w_1}=\dfrac{1+\beta\Delta T}{1+\gamma\Delta T} \). Solving for \( \gamma \) gives the stated result with \( \Delta T=T_2-T_1 \).

Teacher's Note:
1. For thermal expansion use the linear coefficient \( \alpha \) for length, \( 2 \alpha \) for area and \( 3 \alpha \) for volume, valid for small temperature changes.
2. If expansion is prevented, the thermal strain \( \alpha \) times the temperature change equals the mechanical strain, so stress is \( Y \alpha \) times the temperature change.

Answer: (b) Both curves fall exponentially from 60 \(^\circ\)C towards room temperature, and A's curve falls faster, lying below B's curve

Explanation:
1. Same volume and same surface: rate of cooling \( \propto\dfrac{1}{\rho Vc} \). Heat capacities: \( \rho_Ac_A=1.6\times10^6 \), \( \rho_Bc_B=4\times10^6 \).
2. A has the smaller heat capacity, so it cools 2.5 times faster initially: \( A's \) curve lies below \( B's \), both approaching room temperature exponentially.

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.