Thermal Properties of Matter Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 0.25 \, \text{kg} \) iron block at \( 300^\circ \text{C} \) is placed in \( 0.8 \, \text{kg} \) water at \( 24^\circ \text{C} \) in a \( 0.05 \, \text{kg} \) silver calorimeter at \( 24^\circ \text{C} \). What is the final temperature? (Specific heat of iron = \( 450 \, \text{J kg}^{-1} \text{K}^{-1} \), water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), silver = \( 236 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( 0.25 \times 450 \times (300 - T) = (0.8 \times 4186 + 0.05 \times 236) \times (T - 24) \).

\( 33750 - 112.5 T = (3348.8 + 11.8) \times (T - 24) = 3360.6 T - 80654.4 \).

\( 33750 + 80654.4 = 3360.6 T + 112.5 T \).

\( 114404.4 = 3473.1 T \Rightarrow T \approx 32.94^\circ \text{C} \approx 32.9^\circ \text{C} \).

32°C
32.5°C
32.9°C
33°C
3

What happens to the pressure of an ideal gas if its volume is halved and temperature is kept constant?

Boyle’s Law (Section 10.4) states \( P V = \text{constant} \) at constant temperature. If \( V_2 = V_1 / 2 \), then \( P_2 = 2 P_1 \), so the pressure doubles.

It halves
It remains the same
It doubles
It triples
3

Why can’t you open a tightly screwed metal lid until it’s heated in hot water?

Heating the metal lid causes it to expand due to thermal expansion, loosening its grip on the jar, which expands less if made of glass or remains unchanged if plastic.

The lid contracts
The lid expands, loosening it
The jar shrinks
The lid becomes softer
2

A \( 0.1 \, \text{kg} \) lead block at \( 500^\circ \text{C} \) is placed in \( 0.4 \, \text{kg} \) water at \( 26^\circ \text{C} \) in a \( 0.05 \, \text{kg} \) brass calorimeter at \( 26^\circ \text{C} \). What is the final temperature? (Specific heat of lead = \( 127.7 \, \text{J kg}^{-1} \text{K}^{-1} \), water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), brass = \( 386 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( 0.1 \times 127.7 \times (500 - T) = (0.4 \times 4186 + 0.05 \times 386) \times (T - 26) \).

\( 6385 - 12.77 T = (1674.4 + 19.3) \times (T - 26) = 1693.7 T - 44037.2 \).

\( 6385 + 44037.2 = 1693.7 T + 12.77 T \).

\( 50422.2 = 1706.47 T \Rightarrow T \approx 29.55^\circ \text{C} \approx 29.6^\circ \text{C} \).

29°C
29.5°C
30°C
29.6°C
4

Which process describes the direct transition from solid to gas observed in iodine?

Sublimation is the process where a substance transitions directly from solid to gas, as observed with iodine under certain conditions.

Melting
Vaporization
Condensation
Sublimation
4

What happens to the temperature of an ideal gas if its volume doubles at constant pressure?

Charles’ Law states that at constant pressure, \( V / T = \text{constant} \). If volume doubles (\( V_2 = 2V_1 \)), the absolute temperature doubles (\( T_2 = 2T_1 \)).

It halves
It remains the same
It doubles
It triples
3

A \( 0.2 \, \text{kg} \) brass block at \( 350^\circ \text{C} \) is placed in \( 0.5 \, \text{kg} \) water at \( 27^\circ \text{C} \) in a \( 0.1 \, \text{kg} \) iron calorimeter at \( 27^\circ \text{C} \). What is the final temperature? (Specific heat of brass = \( 386 \, \text{J kg}^{-1} \text{K}^{-1} \), water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), iron = \( 450 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( 0.2 \times 386 \times (350 - T) = (0.5 \times 4186 + 0.1 \times 450) \times (T - 27) \).

\( 27020 - 77.2 T = (2093 + 45) \times (T - 27) = 2138 T - 57726 \).

\( 27020 + 57726 = 2138 T + 77.2 T \).

\( 84746 = 2215.2 T \Rightarrow T \approx 38.26^\circ \text{C} \approx 38.3^\circ \text{C} \).

38°C
39°C
38.5°C
38.3°C
4

Why does water in a calorimeter reach a steady temperature when mixed with a hot object?

In calorimetry (Section 10.7), heat lost by the hot object equals heat gained by the water and calorimeter at thermal equilibrium, where no further heat transfer occurs.

Water evaporates
Heat is lost to surroundings
Heat lost equals heat gained
Temperature increases indefinitely
3

A \( 0.25 \, \text{kg} \) silver block at \( 150^\circ \text{C} \) is placed in \( 1 \, \text{kg} \) of water at \( 25^\circ \text{C} \) in a \( 0.2 \, \text{kg} \) aluminium calorimeter at \( 25^\circ \text{C} \). What is the final temperature? (Specific heat of silver = \( 236 \, \text{J kg}^{-1} \text{K}^{-1} \), water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), aluminium = \( 900 \, \text{J kg}^{-1} \text{K}^{-1} \))

Heat lost = Heat gained.

\( 0.25 \times 236 \times (150 - T) = (1 \times 4186 + 0.2 \times 900) \times (T - 25) \).

\( 8850 - 59 T = (4186 + 180) \times (T - 25) = 4366 T - 109150 \).

\( 8850 + 109150 = 4366 T + 59 T \).

\( 118000 = 4425 T \Rightarrow T \approx 26.67^\circ \text{C} \approx 26.7^\circ \text{C} \).

26°C
26.7°C
27°C
28°C
2

A \( 0.4 \, \text{kg} \) aluminium block at \( 160^\circ \text{C} \) is placed in \( 1.2 \, \text{kg} \) water at \( 28^\circ \text{C} \) in a \( 0.2 \, \text{kg} \) lead calorimeter at \( 28^\circ \text{C} \). 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 = \( 127.7 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( 0.4 \times 900 \times (160 - T) = (1.2 \times 4186 + 0.2 \times 127.7) \times (T - 28) \).

\( 57600 - 360 T = (5023.2 + 25.54) \times (T - 28) = 5048.74 T - 141364.72 \).

\( 57600 + 141364.72 = 5048.74 T + 360 T \).

\( 198964.72 = 5408.74 T \Rightarrow T \approx 36.78^\circ \text{C} \approx 36.8^\circ \text{C} \).

36°C
36.8°C
37°C
35°C
2

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