Thermal Properties of Matter — NEET Physics PYQ MCQs with Solutions

Solution: For the increase in length to be independent of temperature, $L\alpha\Delta T$ must be the same for both rods, i.e. $L_{Cu}\alpha_{Cu}=L_{Al}\alpha_{Al}$. With $\alpha_{Cu}=1.7\times10^{-5}$ K$^{-1}$, $\alpha_{Al}=2.2\times10^{-5}$ K$^{-1}$ and $L_{Cu}=88$ cm, $L_{Al}=\dfrac{L_{Cu}\alpha_{Cu}}{\alpha_{Al}}=\dfrac{1.7\times10^{-5}\times88}{2.2\times10^{-5}}\approx68$ cm.

Solution: For the difference $(l_2-l_1)$ to stay the same, the increases must be equal: $l_2'-l_1'=l_2-l_1$, so $l_2(1+\alpha_2\Delta t)-l_1(1+\alpha_1\Delta t)=l_2-l_1$, giving $l_2\alpha_2=l_1\alpha_1$, i.e. $\alpha_1 l_1=\alpha_2 l_2$.

Solution: Since $\rho=\rho_0(1+\gamma\Delta T)$, the fractional change in density is $\dfrac{\rho-\rho_0}{\rho_0}=\gamma\Delta T=5\times10^{-4}\times40=200\times10^{-4}=0.020$.

Solution: Equating the fractional rise on both scales, $\dfrac{39°C-0°C}{100°C-0°C}=\dfrac{t-39°W}{239°W-39°W}$, so $\dfrac{39}{100}=\dfrac{t-39}{200}\Rightarrow t=117°$W.

Solution: For brass, $\Delta l_1=l_1\alpha_1 t$; for steel, $\Delta l_2=l_2\alpha_2 t$. Since $l_2-l_1$ is constant, $\Delta l_2=\Delta l_1$, so $l_2\alpha_2 t=l_1\alpha_1 t$. As $t\neq0$, $l_2\alpha_2=l_1\alpha_1$, i.e. $\alpha_1 l_1=\alpha_2 l_2$.

Solution: A mercury thermometer is a liquid thermometer based on the uniform expansion of mercury. It can be used up to about 300°C or so; mercury begins to vaporise near 367°C. Hence the highest usable value among the options is 360°C.

Solution: Using $\dfrac{C}{100}=\dfrac{F-32}{180}$ with $F=140°$, $\dfrac{C}{100}=\dfrac{140-32}{180}=0.6\Rightarrow C=60°$. Fall in temperature $=100°C-60°C=40°$C.

Solution: Heat required, $Q=mc\Delta T=\left(\dfrac{4}{3}\pi r^3\rho\right)c\Delta T$ (since $m=V_{sphere}\,\rho$). As $\pi,\rho,c$ and $T$ are constants, $Q\propto r^3$, so $\dfrac{Q_1}{Q_2}=\left(\dfrac{r_1}{r_2}\right)^3=(1.5)^3=\dfrac{27}{8}$.

Solution: By conservation of energy, energy gained during the fall from height $h$ is $E=mgh$. Only one-quarter is absorbed by the ice, so $\dfrac{mgh}{4}=mL_f\Rightarrow h=\dfrac{4L_f}{g}=\dfrac{4\times3.4\times10^5}{10}=13.6\times10^4\,\text{m}=136\,\text{km}$.

Solution: Heat lost by the hotter body equals heat gained by the colder body. Because heat capacity increases with temperature, the body at 100°C has a greater heat capacity than the body at 0°C, so the equilibrium temperature is shifted towards 100°C, i.e. $T_c>50°$C (more than 50°C).

Solution: By calorimetry, heat lost by steam = heat gained by water. Let $m'$ be the steam that condenses: $m'L+m's\Delta T=ms\Delta t\Rightarrow m'\times540+m'\times1\times(100-80)=20\times1\times(80-10)\Rightarrow m'=\dfrac{20\times70}{560}=2.5$ g. Net mass of water $=20+2.5=22.5$ g.

Solution: With a constant rate of heat supply, within a single phase $Q=mc\Delta T$ so the temperature rises uniformly, while during a change of state the temperature stays constant, giving a horizontal plateau. The curve that rises, then holds flat during the phase change, then rises again (graph a) represents this behaviour.

Solution: Change in entropy, $\Delta S=\dfrac{ml}{T}=\dfrac{1000\times80}{273}=293$ cal K$^{-1}$.

Solution: Heat to melt 1 g ice at $0^\circ$C: $Q_1=mL=1\times80=80$ cal. Heat to raise 1 g water from $0^\circ$C to $100^\circ$C: $Q_2=mc\Delta\theta=1\times1\times(100-0)=100$ cal. Total $Q=Q_1+Q_2=180$ cal. Heat released by 1 g steam condensing at $100^\circ$C: $Q'=mL'=1\times536=536$ cal. Since $536>180$, not all the steam condenses and the mixture settles at $100^\circ$C.

Solution: Thermal capacity is the heat needed to raise the temperature of the whole body by 1 K, i.e. $=ms=40\times0.2=8$ cal/$^\circ$C $=4.2\times8=33.6$ J/$^\circ$C.

Solution: Vapour pressure of a substance is independent of the amount of substance; it depends only on temperature. Since both containers are at the same temperature, the ratio is $1:1$.

Solution: Let $\theta$ be the equilibrium temperature. Heat gained by ice $=mL+ms(\theta-0)=10\times80+10\times1\times\theta$. Heat lost by the tumbler and its contents $=55\times(40-\theta)$. Equating: $800+10\theta=55(40-\theta)\Rightarrow65\theta=1400\Rightarrow\theta=\dfrac{1400}{65}\approx22^\circ$C.

Solution: By Newton's law of cooling, $\dfrac{\Delta T}{t}=K\left[\dfrac{T_i+T_f}{2}-T_0\right]$. First case: $\dfrac{90-80}{t}=K\left[\dfrac{90+80}{2}-20\right]\Rightarrow\dfrac{10}{t}=65K\Rightarrow K=\dfrac{2}{13t}$. Second case: $\dfrac{80-60}{t_1}=\dfrac{2}{13t}\left[\dfrac{80+60}{2}-20\right]=\dfrac{2}{13t}\times50\Rightarrow t_1=\dfrac{13}{5}t$.

Solution: 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, $T_A > T_C > T_B$.

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

Solution: 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 $d\theta = \rho A\,dx\,L$, $\theta_0 = 0^\circ$C and $\theta_1 = -26^\circ$C. Thus $\dfrac{\rho A\,dx\,L}{dt} = \dfrac{KA(\theta_0 - \theta_1)}{x} \Rightarrow \dfrac{dx}{dt} = \dfrac{K[0-(-26)]}{\rho xL} = \dfrac{26K}{\rho xL}$.

Solution: 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 $P_2 = nP$: $\dfrac{P}{nP} = \dfrac{81}{256} \Rightarrow n = \dfrac{256}{81}$.

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

Solution: 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 $P'=\sigma\left(\dfrac{A}{4}\right)(2T)^4=\dfrac{16}{4}\,\sigma A T^4=4P=4\times450=1800$ watts.

Solution: According to Wien's law, $\lambda_m T=b$, so $\lambda_m=\dfrac{b}{T}=\dfrac{2.88\times10^6\text{ nmK}}{5760\text{ K}}=500$ nm. Since $\lambda_m=500$ nm is the wavelength corresponding to maximum energy, $U_2>U_1$.

Solution: By Newton's law of cooling, $\Delta T=\Delta T_0\,e^{-\lambda t}$. For the first 10 min, $3T-2T=(3T-T)e^{-\lambda\times10}$, giving $e^{-10\lambda}=\dfrac{1}{2}$. Measuring the excess temperature over the full 20 min from the start, $T'-T=(3T-T)e^{-\lambda\times20}=2T\,(e^{-10\lambda})^2=2T\left(\dfrac{1}{2}\right)^2=\dfrac{T}{2}$. Hence $T'=T+\dfrac{T}{2}=\dfrac{3T}{2}$.

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

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

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

Solution: From Stefan's law, $E=\sigma T^4$. The rate of energy production $Q=E\times A=\sigma T^4\times4\pi R^2$. Hence the temperature of the star $T=\left(\dfrac{Q}{4\pi R^2\sigma}\right)^{1/4}$.