Thermal Properties of Matter Chapter-Wise Test 8

Correct answer Carries: 4.

Wrong Answer Carries: -1.

How much ice at \( 0^\circ \text{C} \) will melt if \( 16744 \, \text{J} \) of heat is supplied? (Latent heat of fusion of ice = \( 3.35 \times 10^5 \, \text{J kg}^{-1} \))

Given: \( Q = 16744 \, \text{J} \), \( L_f = 3.35 \times 10^5 \, \text{J kg}^{-1} \).

\( Q = m L_f \Rightarrow m = \frac{Q}{L_f} = \frac{16744}{3.35 \times 10^5} \approx 0.05 \, \text{kg} = 50 \, \text{g} \).

40 g
50 g
60 g
45 g
2

A brass rod of length \( 2.5 \, \text{m} \) at \( 20^\circ \text{C} \) is heated to \( 220^\circ \text{C} \). If its cross-sectional area increases by \( 0.018 \, \text{cm}^2 \), what was its original area? (\( \alpha_l = 1.8 \times 10^{-5} \, \text{K}^{-1} \))

Given: \( \Delta T = 220 - 20 = 200^\circ \text{C} \), \( \Delta A = 0.018 \, \text{cm}^2 \), \( \alpha_l = 1.8 \times 10^{-5} \, \text{K}^{-1} \).

Area expansion: \( \Delta A = A_0 \times 2 \alpha_l \Delta T \).

\( 0.018 = A_0 \times 2 \times 1.8 \times 10^{-5} \times 200 \).

\( 0.018 = A_0 \times 7.2 \times 10^{-3} \Rightarrow A_0 = \frac{0.018}{7.2 \times 10^{-3}} = 2.5 \, \text{cm}^2 \).

2.4 cm²
2.5 cm²
2.6 cm²
2.7 cm²
2

Which property of a substance determines the amount of heat required to change its temperature by \( 1^\circ \text{C} \) per unit mass?

Specific heat capacity (\( s \)) is defined as the heat required per unit mass to change the temperature by \( 1^\circ \text{C} \) (Section 10.6, \( s = \frac{1}{m} \frac{\Delta Q}{\Delta T} \)).

Specific heat capacity
Latent heat
Thermal conductivity
Coefficient of expansion
1

A gas expands from \( 4 \, \text{L} \) to \( 6 \, \text{L} \) at constant pressure of \( 2 \, \text{atm} \) and temperature \( 300 \, \text{K} \). What is the new temperature if the volume is reduced back to \( 4 \, \text{L} \) at constant pressure?

At constant pressure, \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \).

Initial: \( V_1 = 4 \, \text{L} \), \( T_1 = 300 \, \text{K} \), \( V_2 = 6 \, \text{L} \), \( T_2 = ? \).

\( T_2 = T_1 \times \frac{V_2}{V_1} = 300 \times \frac{6}{4} = 450 \, \text{K} \).

Back to \( 4 \, \text{L} \): \( T_3 = T_2 \times \frac{V_3}{V_2} = 450 \times \frac{4}{6} = 300 \, \text{K} \) (original temperature).

290 K
300 K
310 K
450 K
2

A copper cube has a side length of \( 10 \, \text{cm} \) at \( 30^\circ \text{C} \). What is its volume at \( 130^\circ \text{C} \)? (\( \alpha_l = 1.7 \times 10^{-5} \, \text{K}^{-1} \))

Given: \( L_0 = 10 \, \text{cm} \), \( V_0 = 10^3 = 1000 \, \text{cm}^3 \), \( \Delta T = 130 - 30 = 100^\circ \text{C} \), \( \alpha_l = 1.7 \times 10^{-5} \, \text{K}^{-1} \).

Volume expansion: \( \alpha_v = 3 \alpha_l = 3 \times 1.7 \times 10^{-5} = 5.1 \times 10^{-5} \, \text{K}^{-1} \).

\( \Delta V = V_0 \alpha_v \Delta T = 1000 \times 5.1 \times 10^{-5} \times 100 = 5.1 \, \text{cm}^3 \).

\( V = V_0 + \Delta V = 1000 + 5.1 = 1005.1 \, \text{cm}^3 \).

1005.1 cm³
1005 cm³
1006 cm³
1004 cm³
1

Which law explains why the volume of a gas increases when its temperature rises at constant pressure?

Charles’ Law states that at constant pressure, the volume of a gas is directly proportional to its absolute temperature (\( V / T = \text{constant} \), Section 10.4).

Boyle’s Law
Charles’ Law
Ideal Gas Law
Avogadro’s Law
2

A gas at \( 3 \, \text{atm} \) and \( 27^\circ \text{C} \) occupies \( 4 \, \text{L} \). If the pressure is reduced to \( 1.5 \, \text{atm} \) and temperature raised to \( 177^\circ \text{C} \), what is the new volume?

Given: \( P_1 = 3 \, \text{atm} \), \( T_1 = 27^\circ \text{C} = 300 \, \text{K} \), \( V_1 = 4 \, \text{L} \), \( P_2 = 1.5 \, \text{atm} \), \( T_2 = 177^\circ \text{C} = 450 \, \text{K} \).

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

\( V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 4 \times \frac{3}{1.5} \times \frac{450}{300} = 4 \times 2 \times 1.5 = 12 \, \text{L} \).

11 L
11.5 L
13 L
12 L
4

Which gas law relates pressure and volume when temperature is held constant?

Boyle’s Law describes the inverse relationship between pressure and volume of an ideal gas at constant temperature (\( P V = \text{constant} \)).

Charles’ Law
Boyle’s Law
Gay-Lussac’s Law
Avogadro’s Law
2

A \( 0.3 \, \text{kg} \) lead block at \( 400^\circ \text{C} \) is placed in \( 0.7 \, \text{kg} \) water at \( 15^\circ \text{C} \) in a \( 0.1 \, \text{kg} \) aluminium calorimeter at \( 15^\circ \text{C} \). Find the final temperature. (Specific heat of lead = \( 127.7 \, \text{J kg}^{-1} \text{K}^{-1} \), water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), aluminium = \( 900 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( 0.3 \times 127.7 \times (400 - T) = (0.7 \times 4186 + 0.1 \times 900) \times (T - 15) \).

\( 15324 - 38.31 T = (2930.2 + 90) \times (T - 15) = 3020.2 T - 45303 \).

\( 15324 + 45303 = 3020.2 T + 38.31 T \).

\( 60627 = 3058.51 T \Rightarrow T \approx 19.82^\circ \text{C} \approx 19.8^\circ \text{C} \).

19°C
19.5°C
19.8°C
20°C
3

A gas at \( 1.5 \, \text{atm} \) and \( 77^\circ \text{C} \) occupies \( 9 \, \text{L} \). If the pressure is reduced to \( 1 \, \text{atm} \) and temperature lowered to \( 27^\circ \text{C} \), what is the new volume?

Given: \( P_1 = 1.5 \, \text{atm} \), \( T_1 = 77^\circ \text{C} = 350 \, \text{K} \), \( V_1 = 9 \, \text{L} \), \( P_2 = 1 \, \text{atm} \), \( T_2 = 27^\circ \text{C} = 300 \, \text{K} \).

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

\( V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 9 \times \frac{1.5}{1} \times \frac{300}{350} = 9 \times 1.5 \times 0.8571 \approx 11.57 \, \text{L} \).

11.5 L
11.57 L
11.6 L
11.7 L
2

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