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) Jkg-1 (II) JK-1 (III) Jkg-1K^-1 (IV) Jm-1K-1s-1 Choose the correct answer from the options given below:

Governing mathematical equations for thermal properties: 1. Heat capacity: C' = (Delta Q/Delta T) 2. Specific heat capacity: s = (Delta Q/m Delta T) 3. Latent heat: L = (Delta Q/m) 4. Thermal conductivity: K = (Delta Q cdot L/A cdot Delta T cdot t) Let's derive the SI units for each property: - **(A) Heat capacity of body:** - **(B) Specific heat capacity of body:** - **(C) Latent heat:** - **(D) Thermal conductivity:** From heat transfer equation (dQ/dt) = K A (dT/dx): Applying our mappings: - (A) (II) - (B) (III) - (C) (I) - (D) (IV) This maps to Option (4). 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. Class 11 Physics: Thermal Properties of Matter

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) Jkg-1 (II) JK-1 (III) Jkg-1K^-1 (IV) Jm-1K-1s-1 Ch

Governing mathematical equations for thermal properties: 1. Heat capacity: C' = (Delta Q/Delta T) 2. Specific heat capacity: s = (Delta Q/m Delta T) 3. Latent heat: L = (Delta Q/m) 4. Thermal conductivity: K = (Delta Q cdot L/A cdot Delta T cdot t) Let's derive the SI units for each property: - **(A) Heat capacity of body:** - **(B) Specific heat capacity of body:** - **(C) Latent heat:** - **(D) Thermal conductivity:** From heat transfer equation (dQ/dt) = K A (dT/dx): Applying our mappings: -

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) Jkg-1 (II) JK-1 (III) Jkg-1K^-1 (IV) Jm-1K-1s-1 Choose the correct answer from the options given below:

Governing mathematical equations for thermal properties: 1. Heat capacity: C' = (Delta Q/Delta T) 2. Specific heat capacity: s = (Delta Q/m Delta T) 3. Latent heat: L = (Delta Q/m) 4. Thermal conductivity: K = (Delta Q cdot L/A cdot Delta T cdot t) Let's derive the SI units for each property: - **(A) Heat capacity of body:** - **(B) Specific heat capacity of body:** - **(C) Latent heat:** - **(D) Thermal conductivity:** From heat transfer equation (dQ/dt) = K A (dT/dx): Applying our mappings: - (A) (II) - (B) (III) - (C) (I) - (D) (IV) This maps to Option (4). 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. Class 11 Physics: Thermal Properties of Matter

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) Jkg-1 (II) JK-1 (III) Jkg-1K^-1 (IV) Jm-1K-1s-1 Ch

Governing mathematical equations for thermal properties: 1. Heat capacity: C' = (Delta Q/Delta T) 2. Specific heat capacity: s = (Delta Q/m Delta T) 3. Latent heat: L = (Delta Q/m) 4. Thermal conductivity: K = (Delta Q cdot L/A cdot Delta T cdot t) Let's derive the SI units for each property: - **(A) Heat capacity of body:** - **(B) Specific heat capacity of body:** - **(C) Latent heat:** - **(D) Thermal conductivity:** From heat transfer equation (dQ/dt) = K A (dT/dx): Applying our mappings: -

Match List-I with List-II. List-I: (A) Heat capacity of body (B) Specific heat capacity of body (C) Latent heat (D)…

Year	2026	2025	2024	Total
Questions	5	7	1	13

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Governing mathematical equations for thermal properties:

Heat capacity: $C' = (Δ Q)/(Δ T)$ $C' = \frac{\Delta Q}{\Delta T}$
Specific heat capacity: $s = (Δ Q)/(m Δ T)$ $s = \frac{\Delta Q}{m \Delta T}$
Latent heat: $L = (Δ Q)/(m)$ $L = \frac{\Delta Q}{m}$
Thermal conductivity: $K = (Δ Q \cdot L)/(A \cdot Δ T \cdot t)$

Core Logic

Let's derive the SI units for each property:

(A) Heat capacity of body:

[C'] = JK = JK⁻¹ (II)

(B) Specific heat capacity of body:

[s] = Jkg \cdot K = Jkg⁻¹K⁻¹ (III)

(C) Latent heat:

[L] = Jkg = Jkg⁻¹ (I)

(D) Thermal conductivity:

From heat transfer equation $(dQ)/(dt) = K A (dT)/(dx)$ $\frac{dQ}{dt} = K A \frac{dT}{dx}$:

K = (dQ/dt)/(A (dT/dx)) [K] = J/sm² \cdot (K/m) = J \cdot m⁻¹ \cdot K⁻¹ \cdot s⁻¹ (IV)

Step 1: Match values

Applying our mappings:

(A) $\to$ $\to$ (II)
(B) $\to$ $\to$ (III)
(C) $\to$ $\to$ (I)
(D) $\to$ $\to$ (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.

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Class 11 Physics: Thermal Properties of Matter

More Thermal Properties of Matter Previous-Year Questions — Page 3

Q10 jee_main_2025_24_jan_evening Temperature Scales

Which of the following figure represents the relation between Celsius and Fahrenheit temperatures?

A. Graph (1)
B. Graph (2)
C. Graph (3)
D. Graph (4)

Solution

Related Formula

(C)/(5) = (F - 32)/(9)

Core Logic

Rearranging the conversion identity to express $C$ $C$ as a function of $F$ $F$:

C = (5)/(9)F - (160)/(9)

This is a straight line equation $y = mx + c$ $y = mx + c$ with:

Positive slope $m = (5)/(9)$ $m = \frac{5}{9}$
Negative $y$ $y$-intercept $c = -(160)/(9)$ $c = -\frac{160}{9}$ (at $F=0$ $F=0$, $C \approx -17.8°C$ $C \approx -17.8^{\circ}\mathrm{C}$)
Positive $x$ $x$-intercept at $C=0, F=32$ $C=0, F=32$

This completely matches the layout shown in Graph (2).

Correct linear plot for Celsius vs Fahrenheit conversion Q10 — celsius fahrenheit graph, temperature conversion line, linear scaling

Pattern Recognition

$0°C = 32°F$ . Therefore, the line must cross the positive side of the horizontal $F$ $F$ axis when $C=0$ $C=0$.

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Class 11 Physics: Thermal Properties of Matter

Q16 jee_main_2025_24_jan_evening Newton's Law of Cooling

The temperature of a body in air falls from

40°

C to

24°

C in 4 minutes. The temperature of the air is

16°

C. The temperature of the body in the next 4 minutes will be:

A. $(14)/(3)°C$ $\frac{14}{3}^{\circ}\mathrm{C}$
B. $(28)/(3)°C$ $\frac{28}{3}^{\circ}\mathrm{C}$
C. $(56)/(3)°C$ $\frac{56}{3}^{\circ}\mathrm{C}$
D. $(42)/(3)°C$ $\frac{42}{3}^{\circ}\mathrm{C}$

Solution

Related Formula

(T₁ - T₂)/(t) = K[(T₁ + T₂)/(2) - Tₛ]

Core Logic

For the first interval ( $40°C$ $40^{\circ}\mathrm{C}$ to $24°C$ $24^{\circ}\mathrm{C}$ in $4$ $4\ \mathrm{\min}$ with $Tₛ = 16°C$ $T_s = 16^{\circ}\mathrm{C}$):

(40 - 24)/(4) = K[(40 + 24)/(2) - 16]

(16)/(4) = K[32 - 16] 4 = 16K K = (1)/(4)

For the next 4 minutes, let the final temperature be $T$ $T$:

(24 - T)/(4) = (1)/(4)[(24 + T)/(2) - 16]

24 - T = (24 + T)/(2) - 16

40 - T = (24 + T)/(2)

80 - 2T = 24 + T 3T = 56 T = (56)/(3)°C

Pattern Recognition

Newton's law of cooling approximation works beautifully when temperature differences are small. Always compute $K$ $K$ from the first step and substitute directly into the second.

Chapter Mix

Class 11 Physics: Thermal Properties of Matter

Q jee_main_2024_30_january_evening Heating Curve of Water

A block of ice at

-10°C

is slowly heated and converted to steam at

100°C

. Which of the following curves represent the phenomenon qualitatively:

Solution

Core Logic

The heating curve traces temperature vs heat supplied. Stage 1: Ice at $-10°C$ $-10^{\circ}\mathrm{C}$ is heated to $0°C$ $0^{\circ}\mathrm{C}$ (Temperature rises, solid phase). Stage 2: Ice melts at $0°C$ $0^{\circ}\mathrm{C}$ into water (Temperature remains constant until all ice melts). This is a horizontal plateau. Stage 3: Water is heated from $0°C$ $0^{\circ}\mathrm{C}$ to $100°C$ $100^{\circ}\mathrm{C}$ (Temperature rises, liquid phase). Stage 4: Water boils at $100°C$ $100^{\circ}\mathrm{C}$ 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.

Chapter Mix

Class 11 Physics: Thermal Properties of Matter

Solution & Explanation

Related Formula

Core Logic

Step 1: Match values

Pattern Recognition

Chapter Mix

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Related Formula

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