Thermal Properties of Matter Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A silver sheet has an area of \( 1.5 \, \text{m}^2 \) at \( 20^\circ \text{C} \). What is the increase in area when heated to \( 120^\circ \text{C} \)? (\( \alpha_l = 1.9 \times 10^{-5} \, \text{K}^{-1} \))

Given: \( A_0 = 1.5 \, \text{m}^2 \), \( \Delta T = 120 - 20 = 100^\circ \text{C} \), \( \alpha_l = 1.9 \times 10^{-5} \, \text{K}^{-1} \).

\( \Delta A = A_0 \times 2 \alpha_l \Delta T = 1.5 \times 2 \times 1.9 \times 10^{-5} \times 100 = 0.0057 \, \text{m}^2 \).

0.0055 m²
0.0056 m²
0.0058 m²
0.0057 m²
4

Why do blacksmiths heat an iron ring before fitting it onto a wooden wheel?

Heating causes thermal expansion (Section 10.5), increasing the ring’s diameter (\( \Delta L = L_0 \alpha_l \Delta T \)), allowing it to fit over the wheel; it contracts upon cooling for a tight fit (Example 10.2).

To make it softer
To expand it for fitting
To reduce its weight
To increase its strength
2

A gas occupies \( 4.2 \, \text{L} \) at \( 1.4 \, \text{atm} \) and \( 47^\circ \text{C} \). If the pressure is increased by \( 0.6 \, \text{atm} \) and temperature decreased by \( 20^\circ \text{C} \), what is the new volume?

Given: \( V_1 = 4.2 \, \text{L} \), \( P_1 = 1.4 \, \text{atm} \), \( T_1 = 47^\circ \text{C} = 320 \, \text{K} \), \( P_2 = 1.4 + 0.6 = 2 \, \text{atm} \), \( T_2 = 47 - 20 = 27^\circ \text{C} = 300 \, \text{K} \).

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

\( V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 4.2 \times \frac{1.4}{2} \times \frac{300}{320} = 4.2 \times 0.7 \times 0.9375 \approx 2.76 \, \text{L} \).

2.7 L
2.8 L
2.76 L
2.9 L
3

During vaporization, what does the supplied heat primarily do to the liquid?

During vaporization, the supplied heat (latent heat of vaporization) breaks intermolecular bonds to convert the liquid into vapor, without changing its temperature.

Converts it to vapor without temperature change
Increases its temperature
Decreases its density
Expands its volume
1

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

For an ideal gas at constant pressure, Charles’ Law states that volume is directly proportional to absolute temperature. If temperature doubles, volume doubles.

It halves
It doubles
It remains unchanged
It triples
1

What is the Celsius temperature equivalent to \( 400 \, \text{K} \)?

Relation: \( T_C = T - 273.15 \).

Given: \( T = 400 \, \text{K} \).

\( T_C = 400 - 273.15 = 126.85^\circ \text{C} \approx 127^\circ \text{C} \).

126°C
127°C
130°C
125°C
2

A silver rod of length \( 1 \, \text{m} \) at \( 10^\circ \text{C} \) is heated to \( 110^\circ \text{C} \). If its cross-sectional area increases by \( 0.0076 \, \text{cm}^2 \), what was its original area? (\( \alpha_l = 1.9 \times 10^{-5} \, \text{K}^{-1} \))

Given: \( \Delta T = 110 - 10 = 100^\circ \text{C} \), \( \Delta A = 0.0076 \, \text{cm}^2 \), \( \alpha_l = 1.9 \times 10^{-5} \, \text{K}^{-1} \).

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

\( 0.0076 = A_0 \times 2 \times 1.9 \times 10^{-5} \times 100 \).

\( 0.0076 = A_0 \times 3.8 \times 10^{-3} \Rightarrow A_0 = \frac{0.0076}{3.8 \times 10^{-3}} = 2 \, \text{cm}^2 \).

1.9 cm²
1.95 cm²
2.1 cm²
2 cm²
4

A \( 0.15 \, \text{kg} \) copper block at \( 200^\circ \text{C} \) is dropped into \( 0.9 \, \text{kg} \) water at \( 15^\circ \text{C} \) in a calorimeter of mass \( 0.1 \, \text{kg} \) (specific heat = \( 386 \, \text{J kg}^{-1} \text{K}^{-1} \)). 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 = Heat gained.

\( 0.15 \times 386 \times (200 - T) = (0.9 \times 4186 + 0.1 \times 386) \times (T - 15) \).

\( 11580 - 57.9 T = (3767.4 + 38.6) \times (T - 15) = 3806 T - 57090 \).

\( 11580 + 57090 = 3806 T + 57.9 T \).

\( 68670 = 3863.9 T \Rightarrow T \approx 17.77^\circ \text{C} \approx 17.8^\circ \text{C} \).

17°C
17.5°C
17.8°C
18°C
3

A \( 0.05 \, \text{kg} \) gold block at \( 1500^\circ \text{C} \) is placed in \( 0.6 \, \text{kg} \) water at \( 25^\circ \text{C} \) in a \( 0.1 \, \text{kg} \) copper calorimeter at \( 25^\circ \text{C} \). Find the final temperature. (Specific heat of gold = \( 134 \, \text{J kg}^{-1} \text{K}^{-1} \), water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), copper = \( 386 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( 0.05 \times 134 \times (1500 - T) = (0.6 \times 4186 + 0.1 \times 386) \times (T - 25) \).

\( 10050 - 6.7 T = (2511.6 + 38.6) \times (T - 25) = 2550.2 T - 63755 \).

\( 10050 + 63755 = 2550.2 T + 6.7 T \).

\( 73805 = 2556.9 T \Rightarrow T \approx 28.87^\circ \text{C} \approx 28.9^\circ \text{C} \).

28°C
28.5°C
29°C
28.9°C
4

A gas at \( 3.5 \, \text{atm} \) and \( 77^\circ \text{C} \) occupies \( 4.5 \, \text{L} \). If the temperature drops to \( 27^\circ \text{C} \) and volume increases to \( 6 \, \text{L} \), what is the final pressure?

Given: \( P_1 = 3.5 \, \text{atm} \), \( T_1 = 77^\circ \text{C} = 350 \, \text{K} \), \( V_1 = 4.5 \, \text{L} \), \( T_2 = 27^\circ \text{C} = 300 \, \text{K} \), \( V_2 = 6 \, \text{L} \).

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

\( P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 3.5 \times \frac{4.5}{6} \times \frac{300}{350} = 3.5 \times 0.75 \times 0.8571 \approx 2.25 \, \text{atm} \).

2.2 atm
2.3 atm
2.25 atm
2.4 atm
3

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