Thermal Properties of Matter Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A gas at \( 2.4 \, \text{atm} \) and \( 87^\circ \text{C} \) occupies \( 7 \, \text{L} \). If the temperature is lowered to \( 37^\circ \text{C} \) and pressure adjusted to \( 3 \, \text{atm} \), what is the percentage change in volume?

Given: \( P_1 = 2.4 \, \text{atm} \), \( T_1 = 87^\circ \text{C} = 360 \, \text{K} \), \( V_1 = 7 \, \text{L} \), \( T_2 = 37^\circ \text{C} = 310 \, \text{K} \), \( P_2 = 3 \, \text{atm} \).

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

\( V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 7 \times \frac{2.4}{3} \times \frac{310}{360} = 7 \times 0.8 \times 0.8611 \approx 4.82 \, \text{L} \).

Percentage change: \( \frac{V_1 - V_2}{V_1} \times 100 = \frac{7 - 4.82}{7} \times 100 = \frac{2.18}{7} \times 100 \approx 31.14\% \) (decrease).

31.14%
31%
32%
30%
1

How much heat is required to convert \( 0.25 \, \text{kg} \) of ice at \( -30^\circ \text{C} \) to steam at \( 120^\circ \text{C} \)? (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 = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), latent heat of vaporization = \( 2.256 \times 10^6 \, \text{J kg}^{-1} \))

\( Q_1 = 0.25 \times 2100 \times 30 = 15750 \, \text{J} \) (ice to 0°C).

\( Q_2 = 0.25 \times 3.35 \times 10^5 = 83750 \, \text{J} \) (melting).

\( Q_3 = 0.25 \times 4186 \times 100 = 104650 \, \text{J} \) (water to 100°C).

\( Q_4 = 0.25 \times 2.256 \times 10^6 = 564000 \, \text{J} \) (vaporization).

\( Q_5 = 0.25 \times 4186 \times 20 = 20930 \, \text{J} \) (steam to 120°C).

Total: \( Q = 15750 + 83750 + 104650 + 564000 + 20930 = 789080 \, \text{J} = 789.08 \, \text{kJ} \).

785 kJ
788 kJ
790 kJ
789.08 kJ
4

Which of the following correctly relates the coefficients of linear and area expansion?

The coefficient of area expansion is approximately twice the coefficient of linear expansion (\( \Delta A / A \approx 2 \alpha_l \Delta T \)), as area involves two dimensions.

They are equal
Area coefficient is half the linear coefficient
Area coefficient is twice the linear coefficient
Linear coefficient is twice the area coefficient
3

During the melting of ice at \( 0^\circ \text{C} \), why does the temperature remain constant despite heat being supplied?

The heat supplied during melting (latent heat of fusion) is used to change the state from solid to liquid, overcoming intermolecular forces, not to increase temperature (Section 10.8).

Heat is used to change the state, not raise temperature
Ice has a high specific heat
Heat is lost to the surroundings
Temperature decreases during melting
1

How much heat is required to raise \( 0.8 \, \text{kg} \) of aluminium from \( 25^\circ \text{C} \) to \( 75^\circ \text{C} \) if its specific heat capacity is \( 900 \, \text{J kg}^{-1} \text{K}^{-1} \)?

Given: \( m = 0.8 \, \text{kg} \), \( \Delta T = 75 - 25 = 50^\circ \text{C} \), \( s = 900 \, \text{J kg}^{-1} \text{K}^{-1} \).

\( Q = m s \Delta T = 0.8 \times 900 \times 50 = 36000 \, \text{J} = 36 \, \text{kJ} \).

36 kJ
35 kJ
38 kJ
40 kJ
1

How much heat is needed to convert \( 0.1 \, \text{kg} \) of mercury at \( -39^\circ \text{C} \) to liquid at \( 357^\circ \text{C} \)? (Specific heat of solid mercury = \( 140 \, \text{J kg}^{-1} \text{K}^{-1} \), latent heat of fusion = \( 0.12 \times 10^5 \, \text{J kg}^{-1} \), specific heat of liquid mercury = \( 140 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( Q_1 = 0.1 \times 140 \times (39) = 546 \, \text{J} \) (solid from -39°C to 0°C).

\( Q_2 = 0.1 \times 0.12 \times 10^5 = 1200 \, \text{J} \) (melting).

\( Q_3 = 0.1 \times 140 \times 357 = 4998 \, \text{J} \) (liquid from 0°C to 357°C).

Total: \( Q = 546 + 1200 + 4998 = 6744 \, \text{J} = 6.744 \, \text{kJ} \).

6.5 kJ
6.7 kJ
6.744 kJ
7 kJ
3

A copper rod of length \( 80 \, \text{cm} \) at \( 40^\circ \text{C} \) is cooled until its length decreases by \( 0.0136 \, \text{cm} \). What is the final temperature? (\( \alpha_l = 1.7 \times 10^{-5} \, \text{K}^{-1} \))

Given: \( L_0 = 80 \, \text{cm} \), \( \Delta L = -0.0136 \, \text{cm} \), \( \alpha_l = 1.7 \times 10^{-5} \, \text{K}^{-1} \), \( T_1 = 40^\circ \text{C} \).

\( \Delta L = L_0 \alpha_l \Delta T \Rightarrow -0.0136 = 80 \times 1.7 \times 10^{-5} \times \Delta T \).

\( \Delta T = \frac{-0.0136}{80 \times 1.7 \times 10^{-5}} = \frac{-0.0136}{1.36 \times 10^{-3}} = -10 \, \text{K} \).

\( T_2 = 40 - 10 = 30^\circ \text{C} \).

30°C
32°C
28°C
35°C
1

How much heat is required to raise \( 0.4 \, \text{kg} \) of water from \( 15^\circ \text{C} \) to \( 85^\circ \text{C} \) and then convert \( 0.25 \, \text{kg} \) to steam at \( 100^\circ \text{C} \) in a \( 0.2 \, \text{kg} \) silver calorimeter initially at \( 15^\circ \text{C} \)? (Specific heat of water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), silver = \( 236 \, \text{J kg}^{-1} \text{K}^{-1} \), latent heat of vaporization = \( 2.256 \times 10^6 \, \text{J kg}^{-1} \))

\( Q_1 = (0.4 \times 4186 + 0.2 \times 236) \times (85 - 15) = (1674.4 + 47.2) \times 70 = 1721.6 \times 70 = 120512 \, \text{J} \) (to 85°C).

\( Q_2 = (0.4 \times 4186 + 0.2 \times 236) \times (100 - 85) = 1721.6 \times 15 = 25824 \, \text{J} \) (to 100°C).

\( Q_3 = 0.25 \times 2.256 \times 10^6 = 564000 \, \text{J} \) (vaporization).

Total: \( Q = 120512 + 25824 + 564000 = 710336 \, \text{J} = 710.34 \, \text{kJ} \).

710 kJ
712 kJ
710.34 kJ
708 kJ
3

A \( 0.15 \, \text{kg} \) mercury block at \( 400^\circ \text{C} \) is placed in \( 0.5 \, \text{kg} \) water at \( 20^\circ \text{C} \) in a \( 0.05 \, \text{kg} \) aluminium calorimeter at \( 20^\circ \text{C} \). Find the final temperature. (Specific heat of mercury = \( 140 \, \text{J kg}^{-1} \text{K}^{-1} \), water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), aluminium = \( 900 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( 0.15 \times 140 \times (400 - T) = (0.5 \times 4186 + 0.05 \times 900) \times (T - 20) \).

\( 8400 - 21 T = (2093 + 45) \times (T - 20) = 2138 T - 42760 \).

\( 8400 + 42760 = 2138 T + 21 T \).

\( 51160 = 2159 T \Rightarrow T \approx 23.7^\circ \text{C} \).

23°C
23.5°C
24°C
23.7°C
4

Which temperature scale uses absolute zero as its starting point?

The Kelvin scale starts at absolute zero, the theoretical point where all molecular motion ceases, making it an absolute temperature scale.

Kelvin
Celsius
Fahrenheit
Centigrade
1

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