Thermal Properties of Matter JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Thermal Properties of Matter, free to read — no sign-in needed. The full chapter has 29 questions; sign in to attempt the remaining 24 in the exam simulator.

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1. Q1JEE Advanced Adv 2024 (Paper 1)
   The specific heat capacity of a substance is temperature dependent and is given by the formula $C=k T$, where $k$ is a constant of suitable dimensions in SI units, and $T$ is the absolute temperature. If the heat required to raise the temperature of $1 \mathrm{~kg}$ of the substance from $-73^{\circ} \mathrm{C}$ to $27^{\circ} \mathrm{C}$ is $n k$, the value of $n$ is _____ [Given: $0 \mathrm{~K}=-273^{\circ} \mathrm{C}$.]
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   Answer: 25000

   $\begin{aligned} & \mathrm{T}_{\mathrm{i}}=-73^{\circ} \mathrm{C}=200 \mathrm{~K} \\ & \mathrm{~T}_{\mathrm{f}}=27^{\circ} \mathrm{C}=300 \mathrm{~K}\end{aligned}$ $\begin{aligned} & \mathrm{Q}=\int \mathrm{msdT} \\ & =\int 1 \cdot \mathrm{kTdT} \\ & =\int \mathrm{kTdT}=\mathrm{K} \int_{200}^{300} \mathrm{TdT} \\ & =\frac{\mathrm{K}}{2}\left[\mathrm{~T}^2\right]_{200}^{300}=\frac{\mathrm{K}}{2}\left[300^2-200^2\right] \\ & \mathrm{Q}=25000 \mathrm{~K}\end{aligned}$ Hence $n=25000$
2. Q2JEE Advanced Adv 2020 (Paper 2)
   A container with $1kg$ of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is $700W{m}^{-2}$ and it is absorbed by the water over an effective area of $0.05{m}^{2}.$ Assuming that the heat loss from the water to the surroundings is governed by Newton's law of cooling, the difference (in $C^{\circ}$) in the temperature of water and the surroundings after a long time will be_________. (Ignore effect of the container, and take constant for Newton's law of cooling $=0.001{s}^{-1},$ Heat capacity of water $=4200J{kg}^{-1}{K}^{-1}$)
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   Answer: 8.33

   ${P}_{i}=700\times 0.05=35$ watts ${P}_{i}={P}_{r}$ $35=e\sigma A4{T}_{0}^{3}(\Delta T)...\left(1\right)$ $∴K=\dfrac{4e\sigma A{T}_{0}^{3}}{mC}=0.0001$ $4e\sigma A{T}_{0}^{3}=0.001\times mC=4.2$ from equation (1) $\Delta T=\dfrac{35}{4.2}=\dfrac{50}{6}=\dfrac{25}{3}=8.33$
3. Q3JEE Advanced Adv 2020 (Paper 1)
   The filament of a light bulb has surface area $64{mm}^{2}.$ The filament can be considered as a black body at temperature $2500K$ emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of $100m$. Assume the pupil of the eyes of the observer to be circular with radius $3mm$. Then: (Take Stefan-Boltzmann constant $=5.67\times {10}^{-8}W{m}^{-2}{K}^{-4},$ Wiens' displacement constant $=2.90\times {10}^{-3}mK$, Planck's constant $=6.60\times {10}^{-34}Js$, speed of light in vacuum $=3.00\times {10}^{8}m {s}^{-1}$)
   1. A.power radiated by the filament is in the range $642W$ to $645W$
   2. B.radiated power entering into one eye of the observer is in the range $3.15\times {10}^{-8}W$ to $3.25\times {10}^{-8}W$
   3. C.the wavelength corresponding to the maximum intensity of light is $1160nm$
   4. D.taking the average wavelength of emitted radiation to be $1740nm,$ the total number of photons entering per second into one eye of the observer is in the range $2.75\times {10}^{11}$ to $2.85\times {10}^{11}$
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   Answer: B,C,D

   $P=\sigma Ae{T}^{4}$ $P=5.6\times {10}^{-8}\times 64\times {10}^{-6}\times 1\times (2500{)}^{4}$ $P=14175\times {10}^{-14}\times {10}^{8}\times {10}^{4}$ (a) $P=141.75W$ (b) $\dfrac{\sigma Ae{T}^{4}}{4\pi (100{)}^{2}}\times \pi {\left(3\times {10}^{-3}\right)}^{2}=141.75\times \dfrac{9\times {10}^{-6}}{4\times {10}^{4}}$ $318.937\times {10}^{-10}$ $3.18937\times {10}^{8}W$ (c) $\lambda T=b$ $\lambda =\dfrac{2.93\times {10}^{-6}}{2500}=1160nm$ (d) $3.18937\times {10}^{-8}=\left(\dfrac{n}{\sec }\right)\dfrac{hc}{\lambda }$ $3.18937\times \dfrac{{10}^{-8}\lambda }{\lambda e}=n=279.00\times \dfrac{{10}^{-8}\times {10}^{-9}}{{10}^{-34}\times {10}^{8}}$ $n=279\times {10}^{-17}\times {10}^{34}\times {10}^{-8}$ $n=2.79\times {10}^{11}$
4. Q4JEE Advanced Adv 2019 (Paper 1)
   A liquid at ${30}^{o}C$ is poured very slowly into a Calorimeter that is at temperature of ${110}^{o}C.$ The boiling temperature of the liquid is ${80}^{o}C.$ It is found that the first $5gm$ of the liquid completely evaporates. After pouring another $80gm$ of the liquid the equilibrium temperature is found to be ${50}^{o}C.$ The ratio of the Latent heat of the liquid to its specific heat will be ______ ${}^{o}C.$ [Neglect the heat exchange with surrounding]
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   Answer: 270

   Note: The information about condition of calorimeter is not given in question, so following two cases arise - Case $1$ : If calorimeter is closed (vapour not allowed to escape) Heat gain $=$ Heat loss $5S(80-30)+5L=W(110-80)\rightarrow$ (as first $5gm$ liquid is evaporated) $S=$ Specific heat of liquid $L=$ Latent heat of liquid W=Water equivalent of calorimeter $\Rightarrow 250S+5L=W\times 30$ ...(i) Now $80gm$ liquid is poured, Heat gain $=$ Heat loss Here final temperature $=50^{\circ}C$ $∴80\times S\times 20=5L+5S\times 30+W\times 30$ ...(ii) From (i) and (ii) $\dfrac{L}{S}=120 Answer$ Case $2$ : If calorimeter is open and after evaporation liquid escapes $5\times S\times 50+5L=W\times 30$ ...(i) (as first $5gm$ liquid is evaporated) $80\times S\times 20=W\times 30$ ...(ii) (after pouring $80gm$ liquid, the equilibrium temperature is $50^{\circ}C$ $\Rightarrow 80\times S\times 20=S\times S\times 50+5L$ (using (i) and (ii) $SL=1350 S$ $\dfrac{L}{S}=270$
5. Q5JEE Advanced Adv 2019 (Paper 1)
   A current carrying wire heats a metal rod. The wire provides a constant power $\left(P\right)$ to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature $\left(T\right)$ in the metal rod changes with time $\left(t\right)$ as: $T\left(t\right)={T}_{0}\left(1+\beta {t}^{1/4}\right)$ where $\beta$ is a constant with appropriate dimension while ${T}_{0}$ is a constant with dimension of temperature. The heat capacity of the metal is:
   1. A.$\dfrac{4P\left(T\left(t\right)-{T}_{0}\right)}{{\beta }^{4}{T}_{0}^{2}}$
   2. B.$\dfrac{4P{\left(T\left(t\right)-{T}_{0}\right)}^{2}}{{\beta }^{4}{T}_{0}^{3}}$
   3. C.$\dfrac{4P{\left(T\left(t\right)-{T}_{0}\right)}^{4}}{{\beta }^{4}{T}_{0}^{5}}$
   4. D.$\dfrac{4P{\left(T\left(t\right)-{T}_{0}\right)}^{3}}{{\beta }^{4}{T}_{0}^{4}}$
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   Answer: (D)

   Heat required for small increment in temperature of a substance is given by $dQ=m.s.dT$ where, $m=$ mass of substances or $s=$ specific heat of substance Now, heat capacity, $H=m.s.$ $∴ dQ=H.dT$ $\Rightarrow$ Rate of heat absorption by the rod, $\dfrac{dQ}{dt}=P=H\cdot \dfrac{dT}{dt}$ .......(1) But given that, temperature of rod as a function of time is $T={T}_{0}\left(1+\beta {t}^{1/4}\right)$ ......(2) $\Rightarrow \dfrac{dT}{dt}={T}_{0}\cdot \beta \cdot \dfrac{1}{4}{t}^{-\dfrac{3}{4}}$ Putting it in (1) gives $P=H\cdot {T}_{0}\cdot \beta \cdot \dfrac{1}{4}\cdot {t}^{-\dfrac{3}{4}}$ $\Rightarrow H=\dfrac{4P\cdot {t}^{3/4}}{{T}_{0}\cdot \beta }$ $\Rightarrow H=4P\dfrac{{\left(T-{T}_{0}\right)}^{3}}{{T}_{0}^{4}\cdot {B}^{4}}$ Now using equation $(2)$ $T-{T}_{0}={T}_{0}\beta \cdot {t}^{1/4}$ $\Rightarrow {t}^{3/4}={T}_{0}{\left(\dfrac{T-{T}_{0}}{{T}_{0}\beta }\right)}^{3}$