Thermal Properties of Matter Previous Year Questions — JEE Main Physics
13 past-year Thermal Properties of Matter questions from JEE Main (Physics) — page 2.
Q10 (2025)
Which of the following figure represents the relation between Celsius and Fahrenheit temperatures?
- Graph (1)
- Graph (2)
- Graph (3)
- Graph (4)
### Related Formula
$$\frac{C}{5} = \frac{F - 32}{9}$$
### Core Logic
Rearranging the conversion identity to express $C$ as a function of $F$:
$$C = \frac{5}{9}F - \frac{160}{9}$$
This is a straight line equation $y = mx + c$ with:
- Positive slope $m = \frac{5}{9}$
- Negative $y$-intercept $c = -\frac{160}{9}$ (at $F=0$, $C \approx -17.8^{\circ}\mathrm{C}$)
- Positive $x$-intercept at $C=0, F=32$
This completely matches the layout shown in Graph (2). {{SOL_IMG1}}
### Pattern Recognition
$0^{\circ}\mathrm{C} = 32^{\circ}\mathrm{F}$. Therefore, the line must cross the positive side of the horizontal $F$ axis when $C=0$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Q16 (2025)
The temperature of a body in air falls from $40^{\circ}$C to $24^{\circ}$C in 4 minutes. The temperature of the air is $16^{\circ}$C. The temperature of the body in the next 4 minutes will be:
- $\frac{14}{3}^{\circ}\mathrm{C}$
- $\frac{28}{3}^{\circ}\mathrm{C}$
- $\frac{56}{3}^{\circ}\mathrm{C}$
- $\frac{42}{3}^{\circ}\mathrm{C}$
### Related Formula
$$\frac{T_1 - T_2}{t} = K\left[\frac{T_1 + T_2}{2} - T_s\right]$$
### Core Logic
For the first interval ($40^{\circ}\mathrm{C}$ to $24^{\circ}\mathrm{C}$ in $4\ \mathrm{\min}$ with $T_s = 16^{\circ}\mathrm{C}$):
$$\frac{40 - 24}{4} = K\left[\frac{40 + 24}{2} - 16\right]$$
$$\frac{16}{4} = K[32 - 16] \implies 4 = 16K \implies K = \frac{1}{4}$$
For the next 4 minutes, let the final temperature be $T$:
$$\frac{24 - T}{4} = \frac{1}{4}\left[\frac{24 + T}{2} - 16\right]$$
$$24 - T = \frac{24 + T}{2} - 16$$
$$40 - T = \frac{24 + T}{2}$$
$$80 - 2T = 24 + T \implies 3T = 56 \implies T = \frac{56}{3}^{\circ}\mathrm{C}$$
### Pattern Recognition
Newton's law of cooling approximation works beautifully when temperature differences are small. Always compute $K$ from the first step and substitute directly into the second.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Q1714 (2024)
A block of ice at $-10^{\circ}\mathrm{C}$ is slowly heated and converted to steam at $100^{\circ}\mathrm{C}$. Which of the following curves represent the phenomenon qualitatively:
### Core Logic
The heating curve traces temperature vs heat supplied.
Stage 1: Ice at $-10^{\circ}\mathrm{C}$ is heated to $0^{\circ}\mathrm{C}$ (Temperature rises, solid phase).
Stage 2: Ice melts at $0^{\circ}\mathrm{C}$ into water (Temperature remains constant until all ice melts). This is a horizontal plateau.
Stage 3: Water is heated from $0^{\circ}\mathrm{C}$ to $100^{\circ}\mathrm{C}$ (Temperature rises, liquid phase).
Stage 4: Water boils at $100^{\circ}\mathrm{C}$ into steam (Temperature remains constant). This is a second horizontal plateau.
Option (4) correctly shows this sequential step-like graph.
### Pattern Recognition
Heating curves always exhibit horizontal segments during phase changes (latent heat) where temperature remains constant. The slopes of the inclined regions depend on the specific heat capacities of the respective phases.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Q27 (2026)
A gas based geyser heats water flowing at the rate of $5.0$ litres per minute from $27^{\circ}\text{C}$ to $87^{\circ}\text{C}$.
The rate of consumption of the gas is ____ g/s.
(Take heat of combustion of gas = $5.0 \times 10^{4}$ J/g, specific heat capacity of water = $4200\text{ J/kg.}^{\circ}\text{C}$)
- 2.1
- 4.2
- 0.42
- 0.21
### Related Formula
$$P = \frac{dm}{dt} S \Delta T$$
$$P_{\text{heater}} = \text{Rate of combustion} \times \text{Heat of combustion}$$
### Core Logic
Water flow rate = $5\text{ l/min} = \frac{5}{60}\text{ kg/s} = \frac{1}{12}\text{ kg/s}$.
The power required to heat the water is:
$$P_{\text{heater}} = \left(\frac{dm}{dt}\right)_{\text{water}} \cdot S \cdot \Delta T$$
$$P_{\text{heater}} = \frac{1}{12} \times 4200 \times (87 - 27) = \frac{1}{12} \times 4200 \times 60\text{ W}$$
### Step 1: Solving for Gas Consumption
Let the rate of consumption of gas be $x\text{ g/s}$.
Heat generated by the gas per second must equal the power required to heat the water:
$$x \times 5.0 \times 10^{4} = \frac{1}{12} \times 4200 \times 60$$
$$x \times 5.0 \times 10^{4} = 4200 \times 5$$
$$x = \frac{4200 \times 5}{5 \times 10^{4}} = 4200 \times 10^{-4}$$
$$x = 0.42\text{ g/s}$$
{{SOL_IMG1}}
### Pattern Recognition
Equate the thermal power absorbed by fluid ($dm/dt \cdot c \cdot \Delta T$) with the power output of fuel ($dm_{\text{fuel}}/dt \cdot L_{\text{combustion}}$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Q40 (2026)
An aluminium and steel rods having same lengths and cross-sections are joined to make total length of 120 cm at 30°C. The coefficient of linear expansion of aluminium and steel are $24 \times 10^{-6}/^{\circ}\mathrm{C}$ and $1.2 \times 10^{-5}/^{\circ}\mathrm{C}$, respectively. The length of this composite rod when its temperature is raised to 100°C, is ____ cm.
- 120.20
- 120.15
- 120.03
- 120.06
### Related Formula
$$\Delta l = l_0 \alpha \Delta T$$
$$l_{\text{final}} = l_1 + l_2 + \Delta l_1 + \Delta l_2$$
### Core Logic
Let the original lengths be $l_{0A}$ (aluminium) and $l_{0S}$ (steel).
They have same length, so $l_{0A} = l_{0S} = 60\text{ cm}$.
Change in temperature $\Delta T = 100^{\circ}\text{C} - 30^{\circ}\text{C} = 70^{\circ}\text{C}$.
Expansion of Aluminium: $\Delta l_A = l_0 \alpha_A \Delta T$
Expansion of Steel: $\Delta l_S = l_0 \alpha_S \Delta T$
### Step 1: Calculate Final Length
$$l_{\text{final}} = l_0(1 + \alpha_A \Delta T) + l_0(1 + \alpha_S \Delta T)$$
$$l_{\text{final}} = l_0 [2 + (\alpha_A + \alpha_S)\Delta T]$$
$$l_{\text{final}} = 60 \left[ 2 + (24 \times 10^{-6} + 12 \times 10^{-6}) \times 70 \right]$$
$$l_{\text{final}} = 60 \left[ 2 + (36 \times 10^{-6}) \times 70 \right]$$
$$l_{\text{final}} = 60 [2 + 0.00252] = 120 + 60(0.00252) = 120 + 0.1512 = 120.15\text{ cm}$$
{{SOL_IMG1}}
### Pattern Recognition
For composite rods connected end-to-end, expansions simply add up. Convert $1.2 \times 10^{-5}$ to $12 \times 10^{-6}$ immediately for easy mental addition.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
