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

More Thermal Properties of Matter Previous-Year Questions — Page 2

Q24 jee_main_2025_07_april_morning Radiation

A wire of length

10cm

and diameter

0.5mm

is used in a bulb. The temperature of the wire is

1727°C

and power radiated by the wire is 94.2 W. Its emissivity is

x8

where

x =

(Given

σ = 6.0 \times 10⁻⁸ ~W ~m⁻² ~K⁻⁴

pi = 3.14

and assume that the emissivity of wire material is same at all wavelength.)

Numerical Answer. Answer: 5 to 5

Solution

Related Formula

Stefan-Boltzmann Law for radiated thermal power:

P = ε σ A T⁴

Surface area $A$ $A$ of a cylindrical wire of diameter $d$ $d$ and length $L$ $L$ is:

$A = π d L$ $A = \pi d L$

Core Logic

Convert given parameters to standard SI units:

$L = 10 ~cm = 0.1 ~m$ $L = 10 \mathrm{~cm} = 0.1 \mathrm{~m}$
$d = 0.5 ~mm = 0.5 \times 10⁻³ ~m$
$T = 1727 + 273.15 = 2000 ~K$ $T = 1727 + 273.15 = 2000 \mathrm{~K}$
$P = 94.2 ~W$ $P = 94.2 \mathrm{~W}$
$σ = 6.0 \times 10⁻⁸ ~W \cdot m⁻² \cdot K⁻⁴$

Step 1: Express Radiated Power and Emissivity

First, calculate the surface area $A$ $A$:

A = 3.14 \times (0.5 \times 10⁻³) \times 0.1 = 1.57 \times 10⁻⁴ ~m²

Substitute $A$ $A$, $σ$ $\sigma$, and $T$ $T$ into Stefan's formula:

94.2 = ε \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 10¹²

94.2 = ε \times (6 \times 10⁻⁸) \times (1.57 \times 10⁻⁴) \times (16 \times 10¹²)

94.2 = ε \times 6 \times 1.57 \times 16 \times 1 = ε \times 150.72

ε = (94.2)/(150.72) = (5)/(8)

Thus, $(x)/(8) = (5)/(8) x = 5$ .

Pattern Recognition

Sees: Bulb filament heat radiation equation. Shortcut: Simplify the multiplication with factors of 10 first. $T=2000 ~K$ $T=2000 \mathrm{~K}$ has four zeros, so $T⁴$ $T^4$ adds $10¹²$ $10^{12}$ which cancels the $10⁻⁸$ $10^{-8}$ and $10⁻⁴$ $10^{-4}$ from area and constant. The coefficient equation directly yields the ratio $ε = 5/8$ $\epsilon = 5/8$.

Chapter Mix

Class 11 Physics: Thermal Properties of Matter

Q19 jee_main_2025_29_jan_evening Newton's Law of Cooling

A cup of coffee cools from

90°C

80°C

t

minutes when the room temperature is

20°C

. The time taken by the similar cup of coffee to cool from

80°C

60°C

at the same room temperature is :

A. $(13)/(5) t$ $\frac{13}{5} t$
B. $(10)/(13) t$ $\frac{10}{13} t$
C. $(13)/(10) t$ $\frac{13}{10} t$
D. $(5)/(13) t$ $\frac{5}{13} t$

Solution

Related Formula

(Tᵢ - Tf)/(Δ t) = K((Tᵢ + Tf)/(2) - T₀)

Core Logic

Case 1 ( $90°C arrow 80°C$ in time $t$ $t$):

(90 - 80)/(t) = K((90 + 80)/(2) - 20)

(10)/(t) = K(85 - 20) = 65K (i)

Case 2 ( $80°C arrow 60°C$ in time $t'$ $t'$):

(80 - 60)/(t') = K((80 + 60)/(2) - 20)

(20)/(t') = K(70 - 20) = 50K (ii)

Dividing equation $(i)$ $(i)$ by equation $(ii)$ $(ii)$:

(10/t)/(20/t') = (65K)/(50K)

(t')/(2t) = (13)/(10) t' = (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.

Chapter Mix

Class 11 Physics: Thermal Properties of Matter

Q3 jee_main_2025_04_april_evening Thermal Expansion

Consider a rectangular sheet of solid material of length

=9

cm and width

d=4

cm. The coefficient of linear expansion is

α=3.1\times10⁻⁵ K⁻¹

at room temperature and one atmospheric pressure. The mass of sheet

m=0.1 kg

and the specific heat capacity

Cv=900 J kg⁻¹K⁻¹

. If the amount of heat supplied to the material is

8.1\times10²

J then change in area of the rectangular sheet is :-

A. $2.0\times10⁻⁶ m²$ $2.0\times10^{-6}\text{ m}^{2}$
B. $3.0\times10⁻⁷ m²$ $3.0\times10^{-7}\text{ m}^{2}$
C. $6.0\times10⁻⁷ m²$ $6.0\times10^{-7}\text{ m}^{2}$
D. $4.0\times10⁻⁷ m²$ $4.0\times10^{-7}\text{ m}^{2}$

Solution

Related Formula

Δ Q = m Cv Δ T

Δ A = A₀ β Δ T = A₀ (2α) Δ T

Core Logic

First, calculate the temperature change using heat supplied:

Δ T = (Δ Q)/(m Cv)

Substituting the values:

Δ T = (8.1 \times 10²)/(0.1 \times 900) = (810)/(90) = 9 K

Step 1: Calculate Change in Area

Initial area $A₀ = \times d = 9 cm \times 4 cm = 36 cm² = 36 \times 10⁻⁴ m²$ . Now, substitute into the area expansion formula:

Δ A = 36 \times 10⁻⁴ \times 2 \times (3.1 \times 10⁻⁵) \times 9

Δ A = 36 \times 18 \times 3.1 \times 10⁻⁹ = 2008.8 \times 10⁻⁹ m² \approx 2.0 \times 10⁻⁶ m²

Pattern Recognition

Always remember that areal expansion coefficient $β = 2α$ $\beta = 2\alpha$. Convert geometry dimensions to standard units ( $1 cm² = 10⁻⁴ m²$ $1\text{ cm}^2 = 10^{-4}\text{ m}^2$) cleanly before concluding arithmetic.

Chapter Mix

Class 11 Physics: Thermal Properties of Matter

Q23 jee_main_2025_07_april_evening Thermal Conduction

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:

LALB = (1)/(2), rArB = 2

and

KAKB = (1)/(2)

. The free ends of rods A and B are maintained at

400~K

200~K

, respectively. The temperature of rods interface is ________ K, when equilibrium is established. [cite: 187, 188, 189, 190, 191, 192, 193]

Numerical Answer. Answer: 360 to 360

Solution

Related Formula

$Rth = (L)/(KA) = (L)/(K(π r²))$ [cite: 817]

$(dQ)/(dt) = Δ TRth$ [cite: 818]

Core Logic

At steady state equilibrium, the rate of heat flow through both sections in series must be identical: [cite: 193, 819]

$(400 - T)/(R₁) = (T - 200)/(R₂) (400 - T)/(T - 200) = (R₁)/(R₂)$ [cite: 192, 820]

Let's evaluate the resistance ratio $(R₁)/(R₂)$ $\frac{R_1}{R_2}$ using the dimensional parameters: [cite: 820]

$(R₁)/(R₂) = ((LA)/(LB)) \cdot ((rB)/(rA))² \cdot ((KB)/(KA))$ [cite: 820]

Substitute the given values: $(LA)/(LB) = (1)/(2)$ $\frac{L_A}{L_B} = \frac{1}{2}$, $(rA)/(rB) = 2 (rB)/(rA) = (1)/(2)$ , and $(KA)/(KB) = (1)/(2) (KB)/(KA) = 2$ [cite: 189, 190]:

$(R₁)/(R₂) = (1)/(2) \times ((1)/(2))² \times 2 = (1)/(4)$ [cite: 821]

Now link this back into the temperature equation: [cite: 822]

$(400 - T)/(T - 200) = (1)/(4) 1600 - 4T = T - 200$ [cite: 822, 823]

$5T = 1800 T = 360 K$ [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.

Chapter Mix

Class 11 Physics: Thermal Properties of Matter

Solution & Explanation

Related Formula

Core Logic

Step 1: Match values

Pattern Recognition

Chapter Mix

Reference Study Guides

More Thermal Properties of Matter Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Express Radiated Power and Emissivity

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Change in Area

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

More Thermal Properties of Matter Questions — jee_main_2025_02_april_evening

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