NEET 2026-2027 Mock Test: Thermal Properties of Matter

Which temperature scale is based on the absolute minimum temperature where molecular activity theoretically ceases?

The Kelvin scale is defined with its zero point at absolute zero (\(-273.15^\circ \text{C}\)), the theoretical minimum temperature where molecular activity ceases, as per the ideal gas law and thermodynamic principles (Section 10.4).

Kelvin

Celsius

Fahrenheit

Rankine

1

What determines the final temperature when two objects of different temperatures are mixed in a calorimeter?

The final temperature is determined by the principle of conservation of energy, where heat lost by the hotter object equals heat gained by the colder one, reaching thermal equilibrium.

The volume of the objects

The pressure of the system

The material of the calorimeter

Conservation of heat energy

4

What defines the triple point of a substance?

The triple point is the unique temperature and pressure where solid, liquid, and vapor phases coexist in equilibrium (Section 10.8, Triple Point box).

Boiling and freezing occur together

Only vapor exists

Solid, liquid, and vapor coexist

Liquid turns to solid instantly

3

How much heat is required to raise \( 0.6 \, \text{kg} \) of water from \( 25^\circ \text{C} \) to \( 75^\circ \text{C} \) and then convert \( 0.4 \, \text{kg} \) of it to steam at \( 100^\circ \text{C} \) in a \( 0.1 \, \text{kg} \) copper calorimeter initially at \( 25^\circ \text{C} \)? (Specific heat of water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), copper = \( 386 \, \text{J kg}^{-1} \text{K}^{-1} \), latent heat of vaporization = \( 2.256 \times 10^6 \, \text{J kg}^{-1} \))

\( Q_1 = (0.6 \times 4186 + 0.1 \times 386) \times (75 - 25) = (2511.6 + 38.6) \times 50 = 2550.2 \times 50 = 127510 \, \text{J} \) (to 75°C).

\( Q_2 = (0.6 \times 4186 + 0.1 \times 386) \times (100 - 75) = 2550.2 \times 25 = 63755 \, \text{J} \) (to 100°C).

\( Q_3 = 0.4 \times 2.256 \times 10^6 = 902400 \, \text{J} \) (vaporization).

Total: \( Q = 127510 + 63755 + 902400 = 1093665 \, \text{J} = 1093.67 \, \text{kJ} \).

1090 kJ

1092 kJ

1093.67 kJ

1095 kJ

3

A \( 0.1 \, \text{kg} \) iron block at \( 200^\circ \text{C} \) is dropped into \( 0.4 \, \text{kg} \) water at \( 30^\circ \text{C} \) in a \( 0.05 \, \text{kg} \) copper calorimeter at \( 30^\circ \text{C} \). Find the final temperature. (Specific heat of iron = \( 450 \, \text{J kg}^{-1} \text{K}^{-1} \), water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), copper = \( 386 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( 0.1 \times 450 \times (200 - T) = (0.4 \times 4186 + 0.05 \times 386) \times (T - 30) \).

\( 9000 - 45 T = (1674.4 + 19.3) \times (T - 30) = 1693.7 T - 50811 \).

\( 9000 + 50811 = 1693.7 T + 45 T \).

\( 59811 = 1738.7 T \Rightarrow T \approx 34.4^\circ \text{C} \).

34°C

34.4°C

35°C

33°C

2

A steel sphere of radius \( 5 \, \text{cm} \) at \( 20^\circ \text{C} \) is heated to \( 120^\circ \text{C} \). What is the percentage increase in its volume? (\( \alpha_l = 1.2 \times 10^{-5} \, \text{K}^{-1} \))

Given: \( \Delta T = 120 - 20 = 100^\circ \text{C} \), \( \alpha_l = 1.2 \times 10^{-5} \, \text{K}^{-1} \).

\( \alpha_v = 3 \alpha_l = 3 \times 1.2 \times 10^{-5} = 3.6 \times 10^{-5} \, \text{K}^{-1} \).

\( \frac{\Delta V}{V_0} = \alpha_v \Delta T = 3.6 \times 10^{-5} \times 100 = 3.6 \times 10^{-3} \).

Percentage increase: \( \frac{\Delta V}{V_0} \times 100 = 3.6 \times 10^{-3} \times 100 = 0.36\% \).

0.35%

0.36%

0.37%

0.34%

2

A gas occupies \( 3 \, \text{L} \) at \( 47^\circ \text{C} \) and \( 1.5 \, \text{atm} \). What will be its volume if the temperature drops to \( -23^\circ \text{C} \) and pressure increases to \( 3 \, \text{atm} \)?

Given: \( V_1 = 3 \, \text{L} \), \( T_1 = 47^\circ \text{C} = 320 \, \text{K} \), \( P_1 = 1.5 \, \text{atm} \), \( T_2 = -23^\circ \text{C} = 250 \, \text{K} \), \( P_2 = 3 \, \text{atm} \).

\( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \).

\( V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 3 \times \frac{1.5}{3} \times \frac{250}{320} = 3 \times 0.5 \times 0.78125 = 1.171875 \, \text{L} \approx 1.17 \, \text{L} \).

1.2 L

1.17 L

1.3 L

1.1 L

2

Why does a metallic lid expand more than a glass jar when heated together?

Metals typically have a higher coefficient of linear expansion than glass, causing them to expand more for the same temperature increase.

Metal has a higher coefficient of expansion

Glass has higher density

Metal absorbs less heat

Glass expands inward

1

A \( 0.5 \, \text{kg} \) silver block at \( 240^\circ \text{C} \) is placed in \( 1.5 \, \text{kg} \) water at \( 21^\circ \text{C} \) in a \( 0.25 \, \text{kg} \) copper calorimeter at \( 21^\circ \text{C} \). What is the final temperature? (Specific heat of silver = \( 236 \, \text{J kg}^{-1} \text{K}^{-1} \), water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), copper = \( 386 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( 0.5 \times 236 \times (240 - T) = (1.5 \times 4186 + 0.25 \times 386) \times (T - 21) \).

\( 28320 - 118 T = (6279 + 96.5) \times (T - 21) = 6375.5 T - 133885.5 \).

\( 28320 + 133885.5 = 6375.5 T + 118 T \).

\( 162205.5 = 6493.5 T \Rightarrow T \approx 24.97^\circ \text{C} \approx 25^\circ \text{C} \).

24°C

25°C

26°C

23°C

2

A steel rod of length \( 50 \, \text{cm} \) at \( 20^\circ \text{C} \) is heated until its length increases by \( 0.03 \, \text{cm} \). What is the final temperature? (\( \alpha_l = 1.2 \times 10^{-5} \, \text{K}^{-1} \))

Given: \( L_0 = 50 \, \text{cm} \), \( \Delta L = 0.03 \, \text{cm} \), \( \alpha_l = 1.2 \times 10^{-5} \, \text{K}^{-1} \), \( T_1 = 20^\circ \text{C} \).

\( \Delta L = L_0 \alpha_l \Delta T \Rightarrow 0.03 = 50 \times 1.2 \times 10^{-5} \times \Delta T \).

\( \Delta T = \frac{0.03}{50 \times 1.2 \times 10^{-5}} = 500 \, \text{K} \).

\( T_2 = T_1 + \Delta T = 20 + 500 = 520^\circ \text{C} \).

520°C

500°C

540°C

510°C

1

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Thermal Properties Of Matter Chapter-Wise Test 2

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Thermal Properties Of Matter Chapter-Wise Test 2

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