Thermal Properties of Matter JEE Main previous year questions with solutions

5 solved JEE Main questions on Thermal Properties of Matter, free to read — no sign-in needed. The full chapter has 120 questions; sign in to attempt the remaining 115 in the exam simulator.

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1. Q1JEE Main 2026 (05 Apr, Shift 2)Calorimetry
   The heat extracted out of $x$ gram of water initially at $50°C$ to cool it down to $0°C$ is sufficient to evaporate $(1000 - x)$ gram of water also initially at $50°C$. The value of $x$ (closest integer) is _______. (Take latent heat of water $2256\text{ kJ/kg.K}$, specific heat capacity of water $4200\text{ J/kg.K}$)
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   Answer: 922

   Heat extracted to cool $x$ gram of water from $50^{\circ}\text{C}$ to $0^{\circ}\text{C}$ is given by: $Q_1 = m_1 c \Delta T_1$ $Q_1 = x \times 4.2 \times (50 - 0) = 210x \text{ J}$ Heat required to raise the temperature of $(1000 - x)$ gram of water from $50^{\circ}\text{C}$ to $100^{\circ}\text{C}$ and evaporate it is given by: $Q_2 = m_2 c \Delta T_2 + m_2 L$ $Q_2 = (1000 - x) \times [4.2 \times (100 - 50) + 2256]$ $Q_2 = (1000 - x) \times (210 + 2256) = 2466(1000 - x) \text{ J}$ Equating the heat extracted and the heat required: $Q_1 = Q_2$ $210x = 2466(1000 - x)$ $210x = 2466000 - 2466x$ $2676x = 2466000$ $x = \dfrac{2466000}{2676} \approx 921.52$ The closest integer value for $x$ is $922$. Answer: $922$
2. Q2JEE Main 2025 (07 Apr, Shift 1)Radiation
   A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is $1727^{\circ} \mathrm{C}$ and power radiated by the wire is 94.2 W. Its emissivity is $\frac{x}{8}$ where $x=$_______ (Given $\sigma=6.0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}, \pi=3.14$ and assume that the emissivity of wire material is same at all wavelength.)
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   Answer: 5

   $\mathrm{L}=10 \mathrm{~cm}, \mathrm{~d}=0.5 \mathrm{~mm}, \mathrm{~T}=1727^{\circ} \mathrm{C}=2000 \mathrm{~K}$ Power, $\mathrm{P}=94.2 \mathrm{~W}$ $\mathrm{P}=\varepsilon \sigma \mathrm{AT}^4$ $94.2=\varepsilon \times\left(6 \times 10^{-8}\right)(\pi \mathrm{dL})(2000)^4$ $94.2=\varepsilon \times\left(6 \times 10^{-8}\right)(3.14)(0.5)\left(10^{-3}\right)$ $\left(10 \times 10^{-2}\right)(2000)^4$ $\varepsilon=\frac{94.2}{(94.2)(16)}=\frac{5}{8}$
3. Q3JEE Main 2024 (09 Apr, Shift 2)Thermal Expansion
   At room temperature $\left(27^{\circ} \mathrm{C}\right)$, the resistance of a heating element is $50 \Omega$. The temperature coefficient of the material is $2.4 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$. The temperature of the element, when its resistance is $62 \Omega$, is _____ ${ }^{\circ} \mathrm{C}$.
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   Answer: 1027

   $\begin{aligned} & \mathrm{R}=\mathrm{R}_0(1+\alpha \Delta \mathrm{T}) \\ & 62=50\left[1+2.4 \times 10^{-4} \Delta \mathrm{T}\right] \\ & \Delta \mathrm{T}=1000^{\circ} \mathrm{C} \\ & \Rightarrow \mathrm{T}-27^{\circ}=1000^{\circ} \mathrm{C} \\ & \mathrm{T}=1027^{\circ} \mathrm{C}\end{aligned}$
4. Q4JEE Main 2023 (13 Apr, Shift 2)Conduction
   Two plates A and B have thermal conductivities $84W{m}^{-1}{K}^{-1}$ and $126W{m}^{-1}{K}^{-1}$ respectively. They have same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of A and B are kept at ${100}^{^{\circ}}C$ and ${0}^{^{\circ}}C$ respectively, then the temperature of the surface of contact in steady state is_____ $C^{\circ}$.
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   Answer: 40

   Let the temperature of the contact surface be $T$, So, ${H}_{A}={H}_{B}$ $\dfrac{{K}_{A}A'({T}_{A}-T)}{L}=\dfrac{{K}_{B}A'(T-{T}_{B})}{L} \Rightarrow 84(100-T)=126(T-0) \Rightarrow 2(100-T)=3T \Rightarrow T=40^{\circ}C$
5. Q5JEE Main 2021 (01 Sep, Shift 2)Electric Power and Heating Effect of Current
   A steel rod with $y=2.0\times {10}^{11}N{m}^{-2}$ and $\alpha ={10}^{-5}{C^{\circ}}^{-1}$ of length $4m$ and area of cross-section $10{cm}^{2}$ is heated from $0^{\circ}C$ to $400^{\circ}C$ without being allowed to extend. The tension produced in the rod is $x\times {10}^{5}N$ where the value of $x$ is $\underline{\phantom{0000}}.$
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   Answer: 8

   Thermal force $F=Ay\alpha \Delta T$ $F=\left(10\times {10}^{-4}\right)\times \left(2\times {10}^{11}\right)\left({10}^{-5}\right)(400)$ $F=8\times {10}^{5}N$ $\Rightarrow x=8$