Thermal Properties of Matter Previous Year Questions — JEE Main Physics
13 past-year Thermal Properties of Matter questions from JEE Main (Physics).
Q19 (2025)
Match List-I with List-II.
List-I:
(A) Heat capacity of body
(B) Specific heat capacity of body
(C) Latent heat
(D) Thermal conductivity
List-II:
(I) $\mathrm{Jkg^{-1}}$
(II) $\mathrm{JK}^{-1}$
(III) $\mathrm{Jkg^{-1}K^{-1}}$
(IV) $\mathrm{Jm}^{-1}\mathrm{K}^{-1}\mathrm{s}^{-1}$
Choose the correct answer from the options given below:
- $\text{(A)-(III), (B)-(I), (C)-(II), (D)-(IV)}$
- $\text{(A)-(IV), (B)-(III), (C)-(II), (D)-(I)}$
- $\text{(A)-(III), (B)-(IV), (C)-(I), (D)-(II)}$
- $\text{(A)-(II), (B)-(III), (C)-(I), (D)-(IV)}$
### Related Formula
Governing mathematical equations for thermal properties:
1. Heat capacity: $C' = \frac{\Delta Q}{\Delta T}$
2. Specific heat capacity: $s = \frac{\Delta Q}{m \Delta T}$
3. Latent heat: $L = \frac{\Delta Q}{m}$
4. Thermal conductivity: $K = \frac{\Delta Q \cdot L}{A \cdot \Delta T \cdot t}$
### Core Logic
Let's derive the SI units for each property:
- **(A) Heat capacity of body:**
$$[C'] = \frac{\mathrm{J}}{\mathrm{K}} = \mathrm{JK^{-1}} \quad \implies \text{(II)}$$
- **(B) Specific heat capacity of body:**
$$[s] = \frac{\mathrm{J}}{\mathrm{kg \cdot K}} = \mathrm{Jkg^{-1}K^{-1}} \quad \implies \text{(III)}$$
- **(C) Latent heat:**
$$[L] = \frac{\mathrm{J}}{\mathrm{kg}} = \mathrm{Jkg^{-1}} \quad \implies \text{(I)}$$
- **(D) Thermal conductivity:**
From heat transfer equation $\frac{dQ}{dt} = K A \frac{dT}{dx}$:
$$K = \frac{dQ/dt}{A (dT/dx)} \implies [K] = \frac{\mathrm{J/s}}{\mathrm{m}^2 \cdot (\mathrm{K/m})} = \mathrm{J \cdot m^{-1} \cdot K^{-1} \cdot s^{-1}} \quad \implies \text{(IV)}$$
### Step 1: Match values
Applying our mappings:
- (A) $\to$ (II)
- (B) $\to$ (III)
- (C) $\to$ (I)
- (D) $\to$ (IV)
This maps to Option (4).
### Pattern Recognition
Sees: Dimensional unit matching of thermal parameters.
Trap: Confusing Specific Heat Capacity (scaled per mass unit) with Heat Capacity (unscaled entire body parameter).
Shortcut: Specific heat is normalized by mass, meaning its unit must contain kg in the denominator. Heat capacity has no mass constraint, meaning A maps directly to II.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Q24 (2025)
A wire of length $10\mathrm{cm}$ and diameter $0.5\mathrm{mm}$ 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 $\frac{\mathrm{x}}{8}$ where $\mathrm{x} =$
(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.)
### Related Formula
Stefan-Boltzmann Law for radiated thermal power:
$$P = \epsilon \sigma A T^4$$
Surface area $A$ of a cylindrical wire of diameter $d$ and length $L$ is:
$A = \pi d L$
### Core Logic
Convert given parameters to standard SI units:
- $L = 10 \mathrm{~cm} = 0.1 \mathrm{~m}$
- $d = 0.5 \mathrm{~mm} = 0.5 \times 10^{-3} \mathrm{~m}$
- $T = 1727 + 273.15 = 2000 \mathrm{~K}$
- $P = 94.2 \mathrm{~W}$
- $\sigma = 6.0 \times 10^{-8} \mathrm{~W \cdot m^{-2} \cdot K^{-4}}$
### Step 1: Express Radiated Power and Emissivity
First, calculate the surface area $A$:
$$A = 3.14 \times (0.5 \times 10^{-3}) \times 0.1 = 1.57 \times 10^{-4} \mathrm{~m}^2$$
Substitute $A$, $\sigma$, and $T$ into Stefan's formula:
$$94.2 = \epsilon \times (6.0 \times 10^{-8}) \times [3.14 \times (0.5 \times 10^{-3}) \times (10 \times 10^{-2})] \times (2000)^4$$
Calculate temperature term:
$$(2000)^4 = 1.6 \times 10^{12}$$
$$94.2 = \epsilon \times (6 \times 10^{-8}) \times (1.57 \times 10^{-4}) \times (16 \times 10^{12})$$
$$94.2 = \epsilon \times 6 \times 1.57 \times 16 \times 1 = \epsilon \times 150.72$$
$$\epsilon = \frac{94.2}{150.72} = \frac{5}{8}$$
Thus, $\frac{x}{8} = \frac{5}{8} \implies x = 5$.
### Pattern Recognition
Sees: Bulb filament heat radiation equation.
Shortcut: Simplify the multiplication with factors of 10 first. $T=2000 \mathrm{~K}$ has four zeros, so $T^4$ adds $10^{12}$ which cancels the $10^{-8}$ and $10^{-4}$ from area and constant. The coefficient equation directly yields the ratio $\epsilon = 5/8$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Q19 (2025)
A cup of coffee cools from $90^{\circ}\mathrm{C}$ to $80^{\circ}\mathrm{C}$ in $t$ minutes when the room temperature is $20^{\circ}\mathrm{C}$ . The time taken by the similar cup of coffee to cool from $80^{\circ}\mathrm{C}$ to $60^{\circ}\mathrm{C}$ at the same room temperature is :
- $\frac{13}{5} t$
- $\frac{10}{13} t$
- $\frac{13}{10} t$
- $\frac{5}{13} t$
### Related Formula
$$\frac{T_i - T_f}{\Delta t} = K\left(\frac{T_i + T_f}{2} - T_0\right)$$
### Core Logic
**Case 1** ($90^{\circ}\mathrm{C} \rightarrow 80^{\circ}\mathrm{C}$ in time $t$):
$$\frac{90 - 80}{t} = K\left(\frac{90 + 80}{2} - 20
ight)$$
$$\frac{10}{t} = K(85 - 20) = 65K \quad \dots (i)$$
**Case 2** ($80^{\circ}\mathrm{C} \rightarrow 60^{\circ}\mathrm{C}$ in time $t'$):
$$\frac{80 - 60}{t'} = K\left(\frac{80 + 60}{2} - 20
ight)$$
$$\frac{20}{t'} = K(70 - 20) = 50K \quad \dots (ii)$$
Dividing equation $(i)$ by equation $(ii)$:
$$\frac{10/t}{20/t'} = \frac{65K}{50K}$$
$$\frac{t'}{2t} = \frac{13}{10} \implies t' = \frac{13}{5}t$$
### Pattern Recognition
Newton's law of cooling in average form uses the arithmetic mean of temperatures to approximate the driving temperature difference over the interval.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Q3 (2025)
Consider a rectangular sheet of solid material of length $\ell=9$ cm and width $d=4$ cm. The coefficient of linear expansion is $\alpha=3.1\times10^{-5}\text{ K}^{-1}$ at room temperature and one atmospheric pressure. The mass of sheet $m=0.1\text{ kg}$ and the specific heat capacity $C_{v}=900\text{ J kg}^{-1}\text{K}^{-1}$. If the amount of heat supplied to the material is $8.1\times10^{2}$ J then change in area of the rectangular sheet is :-
- $2.0\times10^{-6}\text{ m}^{2}$
- $3.0\times10^{-7}\text{ m}^{2}$
- $6.0\times10^{-7}\text{ m}^{2}$
- $4.0\times10^{-7}\text{ m}^{2}$
### Related Formula
$$\Delta Q = m C_v \Delta T$$
$$\Delta A = A_0 \beta \Delta T = A_0 (2\alpha) \Delta T$$
### Core Logic
First, calculate the temperature change using heat supplied:
$$\Delta T = \frac{\Delta Q}{m C_v}$$
Substituting the values:
$$\Delta T = \frac{8.1 \times 10^2}{0.1 \times 900} = \frac{810}{90} = 9\text{ K}$$
### Step 1: Calculate Change in Area
Initial area $A_0 = \ell \times d = 9\text{ cm} \times 4\text{ cm} = 36\text{ cm}^2 = 36 \times 10^{-4}\text{ m}^2$.
Now, substitute into the area expansion formula:
$$\Delta A = 36 \times 10^{-4} \times 2 \times (3.1 \times 10^{-5}) \times 9$$
$$\Delta A = 36 \times 18 \times 3.1 \times 10^{-9} = 2008.8 \times 10^{-9}\text{ m}^2 \approx 2.0 \times 10^{-6}\text{ m}^2$$
### Pattern Recognition
Always remember that areal expansion coefficient $\beta = 2\alpha$. Convert geometry dimensions to standard units ($1\text{ cm}^2 = 10^{-4}\text{ m}^2$) cleanly before concluding arithmetic.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Q23 (2025)
Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratio of lengths, radii and thermal conductivities of these rods are: $\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}} = \frac{1}{2}, \frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{B}}} = 2$ and $\frac{\mathrm{K}_{\mathrm{A}}}{\mathrm{K}_{\mathrm{B}}} = \frac{1}{2}$ . The free ends of rods A and B are maintained at $400\mathrm{~K}$, $200\mathrm{~K}$, respectively. The temperature of rods interface is ________ K, when equilibrium is established. [cite: 187, 188, 189, 190, 191, 192, 193]
### Related Formula
$$R_{\text{th}} = \frac{L}{KA} = \frac{L}{K(\pi r^2)}$$ [cite: 817]
$$\frac{dQ}{dt} = \frac{\Delta T}{R_{\text{th}}}$$ [cite: 818]
### Core Logic
At steady state equilibrium, the rate of heat flow through both sections in series must be identical: [cite: 193, 819]
$$\frac{400 - T}{R_1} = \frac{T - 200}{R_2} \implies \frac{400 - T}{T - 200} = \frac{R_1}{R_2}$$ [cite: 192, 820]
Let's evaluate the resistance ratio $\frac{R_1}{R_2}$ using the dimensional parameters: [cite: 820]
$$\frac{R_1}{R_2} = \left(\frac{L_A}{L_B}\right) \cdot \left(\frac{r_B}{r_A}\right)^2 \cdot \left(\frac{K_B}{K_A}\right)$$ [cite: 820]
Substitute the given values: $\frac{L_A}{L_B} = \frac{1}{2}$, $\frac{r_A}{r_B} = 2 \implies \frac{r_B}{r_A} = \frac{1}{2}$, and $\frac{K_A}{K_B} = \frac{1}{2} \implies \frac{K_B}{K_A} = 2$ [cite: 189, 190]:
$$\frac{R_1}{R_2} = \frac{1}{2} \times \left(\frac{1}{2}\right)^2 \times 2 = \frac{1}{4}$$ [cite: 821]
Now link this back into the temperature equation: [cite: 822]
$$\frac{400 - T}{T - 200} = \frac{1}{4} \implies 1600 - 4T = T - 200$$ [cite: 822, 823]
$$5T = 1800 \implies T = 360\ \text{K}$$ [cite: 824, 825]
### Pattern Recognition
Thermal conduction processes behave exactly like electric current fields in series connections[cite: 818, 819]. Cross-sectional area scales squarely with radius parameters, which requires extra care during substitution.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
