Thermal Properties of Matter Chapter-Wise Test 10

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A copper block of mass \( 0.5 \, \text{kg} \) at \( 100^\circ \text{C} \) is dropped into \( 1 \, \text{kg} \) of water at \( 20^\circ \text{C} \). Find the final temperature. (Specific heat of copper = \( 386 \, \text{J kg}^{-1} \text{K}^{-1} \), water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \))

Heat lost by copper = Heat gained by water.

\( m_c s_c (100 - T) = m_w s_w (T - 20) \).

\( 0.5 \times 386 \times (100 - T) = 1 \times 4186 \times (T - 20) \).

\( 19300 - 193 T = 4186 T - 83720 \).

\( 19300 + 83720 = 4186 T + 193 T \).

\( 103020 = 4379 T \Rightarrow T \approx 23.53^\circ \text{C} \).

22°C
23.5°C
25°C
20°C
2

A brass rod of length \( 2 \, \text{m} \) at \( 25^\circ \text{C} \) is heated to \( 125^\circ \text{C} \). Calculate the increase in length. (Coefficient of linear expansion of brass = \( 1.8 \times 10^{-5} \, \text{K}^{-1} \))

Given: \( L_0 = 2 \, \text{m} \), \( \Delta T = 125 - 25 = 100^\circ \text{C} \), \( \alpha_l = 1.8 \times 10^{-5} \, \text{K}^{-1} \).

\( \Delta L = L_0 \alpha_l \Delta T = 2 \times 1.8 \times 10^{-5} \times 100 = 3.6 \times 10^{-3} \, \text{m} = 3.6 \, \text{mm} \).

3.5 mm
3.6 mm
3.8 mm
4.0 mm
2

A silver wire of length \( 1.2 \, \text{m} \) at \( 25^\circ \text{C} \) is cooled until its length decreases by \( 0.0228 \, \text{cm} \). What is the final temperature? (\( \alpha_l = 1.9 \times 10^{-5} \, \text{K}^{-1} \))

Given: \( L_0 = 1.2 \, \text{m} = 120 \, \text{cm} \), \( \Delta L = -0.0228 \, \text{cm} \), \( \alpha_l = 1.9 \times 10^{-5} \, \text{K}^{-1} \), \( T_1 = 25^\circ \text{C} \).

\( \Delta L = L_0 \alpha_l \Delta T \Rightarrow -0.0228 = 120 \times 1.9 \times 10^{-5} \times \Delta T \).

\( \Delta T = \frac{-0.0228}{120 \times 1.9 \times 10^{-5}} = \frac{-0.0228}{2.28 \times 10^{-3}} = -10 \, \text{K} \).

\( T_2 = 25 - 10 = 15^\circ \text{C} \).

15°C
16°C
14°C
17°C
1

A \( 0.15 \, \text{kg} \) aluminium block at \( 280^\circ \text{C} \) is placed in \( 0.7 \, \text{kg} \) water at \( 22^\circ \text{C} \) in a \( 0.05 \, \text{kg} \) lead calorimeter at \( 22^\circ \text{C} \). What is the final temperature? (Specific heat of aluminium = \( 900 \, \text{J kg}^{-1} \text{K}^{-1} \), water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), lead = \( 127.7 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( 0.15 \times 900 \times (280 - T) = (0.7 \times 4186 + 0.05 \times 127.7) \times (T - 22) \).

\( 37800 - 135 T = (2930.2 + 6.385) \times (T - 22) = 2936.585 T - 64599.87 \).

\( 37800 + 64599.87 = 2936.585 T + 135 T \).

\( 102399.87 = 3071.585 T \Rightarrow T \approx 33.34^\circ \text{C} \approx 33.3^\circ \text{C} \).

33°C
34°C
33.3°C
32°C
3

Why do gases expand more than solids for the same temperature increase?

Gases have a much higher coefficient of volume expansion than solids due to weaker intermolecular forces, allowing greater volume changes with temperature.

Gases have higher density
Solids have higher specific heat
Gases have lower pressure
Gases have a higher expansion coefficient
4

A brass ring has an inner diameter of \( 10 \, \text{cm} \) at \( 40^\circ \text{C} \). What temperature must it be cooled to for the diameter to decrease by \( 0.009 \, \text{cm} \)? (\( \alpha_l = 1.8 \times 10^{-5} \, \text{K}^{-1} \))

Given: \( L_0 = 10 \, \text{cm} \), \( \Delta L = -0.009 \, \text{cm} \), \( \alpha_l = 1.8 \times 10^{-5} \, \text{K}^{-1} \), \( T_1 = 40^\circ \text{C} \).

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

\( \Delta T = \frac{-0.009}{10 \times 1.8 \times 10^{-5}} = \frac{-0.009}{1.8 \times 10^{-4}} = -50 \, \text{K} \).

\( T_2 = 40 - 50 = -10^\circ \text{C} \).

-8°C
-10°C
-12°C
-9°C
2

How much heat is required to raise the temperature of \( 2 \, \text{kg} \) of water from \( 20^\circ \text{C} \) to \( 50^\circ \text{C} \)? (Specific heat capacity of water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \))

Given: \( m = 2 \, \text{kg} \), \( \Delta T = 50 - 20 = 30^\circ \text{C} \), \( s = 4186 \, \text{J kg}^{-1} \text{K}^{-1} \).

Heat required: \( \Delta Q = m s \Delta T = 2 \times 4186 \times 30 = 251160 \, \text{J} \).

250 kJ
251.16 kJ
260 kJ
240 kJ
2

Which temperature scale uses \( 32^\circ \) and \( 212^\circ \) as the freezing and boiling points of water at standard pressure?

The Fahrenheit scale defines the freezing point of water as \( 32^\circ \text{F} \) and boiling point as \( 212^\circ \text{F} \) at standard pressure (Section 10.3).

Celsius
Kelvin
Fahrenheit
Rankine
3

What happens to the boiling point of a liquid when external pressure decreases?

Decreasing external pressure reduces the energy needed for molecules to escape into the vapor phase, lowering the boiling point, as seen in high-altitude cooking.

It increases
It decreases
It remains constant
It doubles
2

A silver ring has an inner circumference of \( 31.4 \, \text{cm} \) at \( 15^\circ \text{C} \). What temperature must it be heated to for the circumference to increase by \( 0.0597 \, \text{cm} \)? (\( \alpha_l = 1.9 \times 10^{-5} \, \text{K}^{-1} \))

Given: \( L_0 = 31.4 \, \text{cm} \), \( \Delta L = 0.0597 \, \text{cm} \), \( \alpha_l = 1.9 \times 10^{-5} \, \text{K}^{-1} \), \( T_1 = 15^\circ \text{C} \).

\( \Delta L = L_0 \alpha_l \Delta T \Rightarrow 0.0597 = 31.4 \times 1.9 \times 10^{-5} \times \Delta T \).

\( \Delta T = \frac{0.0597}{31.4 \times 1.9 \times 10^{-5}} = \frac{0.0597}{5.966 \times 10^{-4}} \approx 100 \, \text{K} \).

\( T_2 = 15 + 100 = 115^\circ \text{C} \).

110°C
112°C
115°C
120°C
3

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