NEET 2026-2027 Mock Test: Thermal Properties of Matter

Which substance typically has the highest specific heat capacity among common materials?

Water has the highest specific heat capacity (\( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \)) among common substances listed (Section 10.6, Table 10.3), making it an effective coolant.

Aluminium

Water

Copper

Iron

2

A gas at \( 37^\circ \text{C} \) and \( 3 \, \text{atm} \) occupies \( 6 \, \text{L} \). If it is cooled to \( -13^\circ \text{C} \) while the volume is reduced to \( 4 \, \text{L} \), what is the final pressure?

Given: \( T_1 = 37^\circ \text{C} = 310 \, \text{K} \), \( P_1 = 3 \, \text{atm} \), \( V_1 = 6 \, \text{L} \), \( T_2 = -13^\circ \text{C} = 260 \, \text{K} \), \( V_2 = 4 \, \text{L} \).

Ideal gas law: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \).

\( P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 3 \times \frac{6}{4} \times \frac{260}{310} \).

\( P_2 = 3 \times 1.5 \times 0.8387 \approx 3.77 \, \text{atm} \).

3.77 atm

3.8 atm

4.0 atm

3.5 atm

1

A gas at \( 1.5 \, \text{atm} \) and \( 47^\circ \text{C} \) occupies \( 8 \, \text{L} \). If the temperature is raised to \( 97^\circ \text{C} \) and pressure adjusted to \( 2 \, \text{atm} \), what is the new volume?

Given: \( P_1 = 1.5 \, \text{atm} \), \( T_1 = 47^\circ \text{C} = 320 \, \text{K} \), \( V_1 = 8 \, \text{L} \), \( T_2 = 97^\circ \text{C} = 370 \, \text{K} \), \( P_2 = 2 \, \text{atm} \).

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

\( V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 8 \times \frac{1.5}{2} \times \frac{370}{320} \).

\( V_2 = 8 \times 0.75 \times 1.15625 = 6.9375 \, \text{L} \approx 6.94 \, \text{L} \).

6.9 L

6.94 L

7.0 L

6.8 L

2

A \( 0.3 \, \text{kg} \) lead block at \( 350^\circ \text{C} \) is placed in \( 0.6 \, \text{kg} \) water at \( 25^\circ \text{C} \). Find the final temperature. (Specific heat of lead = \( 127.7 \, \text{J kg}^{-1} \text{K}^{-1} \), water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( 0.3 \times 127.7 \times (350 - T) = 0.6 \times 4186 \times (T - 25) \).

\( 13408.5 - 38.31 T = 2511.6 T - 62790 \).

\( 13408.5 + 62790 = 2511.6 T + 38.31 T \).

\( 76198.5 = 2549.91 T \Rightarrow T \approx 29.89^\circ \text{C} \approx 29.9^\circ \text{C} \).

29°C

29.9°C

30°C

31°C

2

A copper plate has an area of \( 0.8 \, \text{m}^2 \) at \( 60^\circ \text{C} \). What is the decrease in area when cooled to \( 10^\circ \text{C} \)? (\( \alpha_l = 1.7 \times 10^{-5} \, \text{K}^{-1} \))

Given: \( A_0 = 0.8 \, \text{m}^2 \), \( \Delta T = 10 - 60 = -50^\circ \text{C} \), \( \alpha_l = 1.7 \times 10^{-5} \, \text{K}^{-1} \).

\( \Delta A = A_0 \times 2 \alpha_l \Delta T = 0.8 \times 2 \times 1.7 \times 10^{-5} \times (-50) \).

\( \Delta A = 0.8 \times 3.4 \times 10^{-5} \times (-50) = -0.00136 \, \text{m}^2 \) (decrease of \( 0.00136 \, \text{m}^2 \)).

0.0013 m²

0.0014 m²

0.00136 m²

0.00135 m²

3

Which of the following is true about the coefficient of volume expansion compared to the coefficient of linear expansion?

The coefficient of volume expansion (\( \alpha_v \)) is three times the coefficient of linear expansion (\( \alpha_l \)), as \( \alpha_v = 3 \alpha_l \) (Section 10.5).

They are equal

\( \alpha_v = 2 \alpha_l \)

\( \alpha_v = 3 \alpha_l \)

\( \alpha_l = 3 \alpha_v \)

3

A steel cylinder has a volume of \( 1.5 \, \text{L} \) at \( 55^\circ \text{C} \). What temperature must it be heated to for the volume to increase by \( 0.0054 \, \text{L} \)? (\( \alpha_l = 1.2 \times 10^{-5} \, \text{K}^{-1} \))

Given: \( V_0 = 1.5 \, \text{L} = 1500 \, \text{cm}^3 \), \( \Delta V = 0.0054 \, \text{L} = 5.4 \, \text{cm}^3 \), \( \alpha_l = 1.2 \times 10^{-5} \, \text{K}^{-1} \), \( T_1 = 55^\circ \text{C} \).

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

\( \Delta V = V_0 \alpha_v \Delta T \Rightarrow 5.4 = 1500 \times 3.6 \times 10^{-5} \times \Delta T \).

\( \Delta T = \frac{5.4}{1500 \times 3.6 \times 10^{-5}} = \frac{5.4}{0.054} = 100 \, \text{K} \).

\( T_2 = 55 + 100 = 155^\circ \text{C} \).

150°C

152°C

160°C

155°C

4

What is the Fahrenheit temperature equivalent to \( 75^\circ \text{C} \)?

\( t_F = \frac{9}{5} t_C + 32 = \frac{9}{5} \times 75 + 32 = 135 + 32 = 167^\circ \text{F} \).

165°F

167°F

170°F

160°F

2

What happens to the volume of an ideal gas if its pressure is doubled while the temperature remains constant?

According to Boyle’s Law (Section 10.4), for an ideal gas at constant temperature, \( P V = \text{constant} \). If pressure doubles (\( P_2 = 2P_1 \)), then \( V_2 = V_1 / 2 \), so the volume halves.

It doubles

It halves

It remains the same

It triples

1

How much heat is required to convert \( 0.4 \, \text{kg} \) of ice at \( -25^\circ \text{C} \) to water at \( 50^\circ \text{C} \)? (Specific heat of ice = \( 2100 \, \text{J kg}^{-1} \text{K}^{-1} \), latent heat of fusion = \( 3.35 \times 10^5 \, \text{J kg}^{-1} \), specific heat of water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( Q_1 = 0.4 \times 2100 \times 25 = 21000 \, \text{J} \) (ice to 0°C).

\( Q_2 = 0.4 \times 3.35 \times 10^5 = 134000 \, \text{J} \) (melting).

\( Q_3 = 0.4 \times 4186 \times 50 = 83720 \, \text{J} \) (water to 50°C).

Total: \( Q = 21000 + 134000 + 83720 = 238720 \, \text{J} = 238.72 \, \text{kJ} \).

235 kJ

238 kJ

238.72 kJ

240 kJ

3

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Thermal Properties Of Matter Chapter-Wise Test 3

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Thermal Properties Of Matter Chapter-Wise Test 3

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