Thermal Properties of Matter Chapter-Wise Test 6

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A brass ring has an inner diameter of \( 5 \, \text{cm} \) at \( 30^\circ \text{C} \). To what temperature must it be heated to increase the diameter to \( 5.009 \, \text{cm} \)? (\( \alpha_l = 1.8 \times 10^{-5} \, \text{K}^{-1} \))

Given: \( L_0 = 5 \, \text{cm} \), \( L = 5.009 \, \text{cm} \), \( \alpha_l = 1.8 \times 10^{-5} \, \text{K}^{-1} \), \( T_1 = 30^\circ \text{C} \).

\( \Delta L = 5.009 - 5 = 0.009 \, \text{cm} \).

\( \Delta L = L_0 \alpha_l \Delta T \Rightarrow 0.009 = 5 \times 1.8 \times 10^{-5} \times \Delta T \).

\( \Delta T = \frac{0.009}{5 \times 1.8 \times 10^{-5}} = 100 \, \text{K} \).

\( T_2 = 30 + 100 = 130^\circ \text{C} \).

120°C
125°C
130°C
135°C
3

How much heat is required to vaporize \( 0.15 \, \text{kg} \) of nitrogen at \( -196^\circ \text{C} \)? (Latent heat of vaporization of nitrogen = \( 2.0 \times 10^5 \, \text{J kg}^{-1} \))

Given: \( m = 0.15 \, \text{kg} \), \( L_v = 2.0 \times 10^5 \, \text{J kg}^{-1} \).

\( Q = m L_v = 0.15 \times 2.0 \times 10^5 = 30000 \, \text{J} = 30 \, \text{kJ} \).

28 kJ
29 kJ
31 kJ
30 kJ
4

How much heat is required to convert \( 0.3 \, \text{kg} \) of ice at \( -18^\circ \text{C} \) to water at \( 40^\circ \text{C} \) in a \( 0.1 \, \text{kg} \) copper calorimeter initially at \( 25^\circ \text{C} \)? (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 = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), copper = \( 386 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( Q_1 = 0.3 \times 2100 \times 18 = 11340 \, \text{J} \) (ice to 0°C).

\( Q_2 = 0.3 \times 3.35 \times 10^5 = 100500 \, \text{J} \) (melting).

Heat gained by water and calorimeter: \( (0.3 \times 4186 + 0.1 \times 386) \times (40 - 0) = (1255.8 + 38.6) \times 40 = 1294.4 \times 40 = 51776 \, \text{J} \).

Heat lost by calorimeter: \( 0.1 \times 386 \times (25 - 0) = 965 \, \text{J} \) (assume it cools to 0°C first).

Total heat supplied: \( Q = 11340 + 100500 + 51776 - 965 = 162651 \, \text{J} = 162.65 \, \text{kJ} \).

162 kJ
162.65 kJ
163 kJ
161 kJ
2

A \( 0.05 \, \text{kg} \) gold block at \( 1200^\circ \text{C} \) is placed in \( 0.5 \, \text{kg} \) water at \( 20^\circ \text{C} \). What is the final temperature? (Specific heat of gold = \( 134 \, \text{J kg}^{-1} \text{K}^{-1} \), water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( 0.05 \times 134 \times (1200 - T) = 0.5 \times 4186 \times (T - 20) \).

\( 8040 - 6.7 T = 2093 T - 41860 \).

\( 8040 + 41860 = 2093 T + 6.7 T \).

\( 49900 = 2099.7 T \Rightarrow T \approx 23.77^\circ \text{C} \approx 23.8^\circ \text{C} \).

23°C
23.5°C
24°C
23.8°C
4

What is the significance of the latent heat of fusion?

The latent heat of fusion is the heat required per unit mass to change a substance from solid to liquid at its melting point, reflecting the energy needed to break intermolecular bonds.

It raises the temperature
It cools the substance
It changes solid to liquid
It changes liquid to vapor
3

A \( 0.1 \, \text{kg} \) aluminium block at \( 80^\circ \text{C} \) is placed in \( 0.5 \, \text{kg} \) water at \( 25^\circ \text{C} \). Calculate the final temperature. (Specific heat of aluminium = \( 900 \, \text{J kg}^{-1} \text{K}^{-1} \), water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \))

Heat lost = Heat gained.

\( 0.1 \times 900 \times (80 - T) = 0.5 \times 4186 \times (T - 25) \).

\( 7200 - 90 T = 2093 T - 52325 \).

\( 7200 + 52325 = 2093 T + 90 T \).

\( 59525 = 2183 T \Rightarrow T \approx 27.27^\circ \text{C} \).

26°C
27.3°C
28°C
25°C
2

A gas occupies a volume of \( 2 \, \text{L} \) at \( 27^\circ \text{C} \) and \( 1 \, \text{atm} \) pressure. What will be its volume at \( 127^\circ \text{C} \) if the pressure remains constant? (Use ideal gas law)

Given: \( V_1 = 2 \, \text{L} \), \( T_1 = 27^\circ \text{C} = 300 \, \text{K} \), \( T_2 = 127^\circ \text{C} = 400 \, \text{K} \), \( P_1 = P_2 \).

Using Charles' Law: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \).

\( V_2 = V_1 \times \frac{T_2}{T_1} = 2 \times \frac{400}{300} = 2 \times \frac{4}{3} = \frac{8}{3} \approx 2.67 \, \text{L} \).

2.5 L
2.67 L
3.0 L
2.0 L
2

A metal cube has a volume of \( 1000 \, \text{cm}^3 \) at \( 0^\circ \text{C} \). What will be its volume at \( 100^\circ \text{C} \) if the coefficient of volume expansion is \( 5.4 \times 10^{-5} \, \text{K}^{-1} \)?

Given: \( V_0 = 1000 \, \text{cm}^3 \), \( \Delta T = 100^\circ \text{C} \), \( \alpha_v = 5.4 \times 10^{-5} \, \text{K}^{-1} \).

\( \Delta V = V_0 \alpha_v \Delta T = 1000 \times 5.4 \times 10^{-5} \times 100 = 5.4 \, \text{cm}^3 \).

New volume: \( V = V_0 + \Delta V = 1000 + 5.4 = 1005.4 \, \text{cm}^3 \).

1004 cm³
1005.4 cm³
1006 cm³
1005 cm³
2

A \( 0.2 \, \text{kg} \) tungsten block at \( 250^\circ \text{C} \) is dropped into \( 0.5 \, \text{kg} \) water at \( 18^\circ \text{C} \) in a \( 0.1 \, \text{kg} \) silver calorimeter at \( 18^\circ \text{C} \). What is the final temperature? (Specific heat of tungsten = \( 134 \, \text{J kg}^{-1} \text{K}^{-1} \), water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), silver = \( 236 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( 0.2 \times 134 \times (250 - T) = (0.5 \times 4186 + 0.1 \times 236) \times (T - 18) \).

\( 6700 - 26.8 T = (2093 + 23.6) \times (T - 18) = 2116.6 T - 38098.8 \).

\( 6700 + 38098.8 = 2116.6 T + 26.8 T \).

\( 44798.8 = 2143.4 T \Rightarrow T \approx 20.9^\circ \text{C} \).

20.9°C
21°C
20°C
22°C
1

A gas at \( 17^\circ \text{C} \) and \( 2.5 \, \text{atm} \) occupies \( 5 \, \text{L} \). If the temperature is raised to \( 67^\circ \text{C} \) and the volume is adjusted to \( 7 \, \text{L} \), what is the final pressure?

Given: \( T_1 = 17^\circ \text{C} = 290 \, \text{K} \), \( P_1 = 2.5 \, \text{atm} \), \( V_1 = 5 \, \text{L} \), \( T_2 = 67^\circ \text{C} = 340 \, \text{K} \), \( V_2 = 7 \, \text{L} \).

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

\( P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 2.5 \times \frac{5}{7} \times \frac{340}{290} \).

\( P_2 = 2.5 \times 0.7143 \times 1.1724 \approx 2.09 \, \text{atm} \).

2.09 atm
2.1 atm
2.2 atm
2.0 atm
1

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