Thermal Properties of Matter Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A copper cylinder has a volume of \( 500 \, \text{cm}^3 \) at \( 50^\circ \text{C} \). What temperature must it be cooled to for the volume to decrease by \( 0.85 \, \text{cm}^3 \)? (\( \alpha_l = 1.7 \times 10^{-5} \, \text{K}^{-1} \))

Given: \( V_0 = 500 \, \text{cm}^3 \), \( \Delta V = -0.85 \, \text{cm}^3 \), \( \alpha_l = 1.7 \times 10^{-5} \, \text{K}^{-1} \), \( T_1 = 50^\circ \text{C} \).

\( \alpha_v = 3 \alpha_l = 3 \times 1.7 \times 10^{-5} = 5.1 \times 10^{-5} \, \text{K}^{-1} \).

\( \Delta V = V_0 \alpha_v \Delta T \Rightarrow -0.85 = 500 \times 5.1 \times 10^{-5} \times \Delta T \).

\( \Delta T = \frac{-0.85}{500 \times 5.1 \times 10^{-5}} = \frac{-0.85}{0.0255} \approx -33.33 \, \text{K} \).

\( T_2 = 50 - 33.33 \approx 16.67^\circ \text{C} \approx 16.7^\circ \text{C} \).

16°C
17°C
16.5°C
16.7°C
4

Which scale measures temperature relative to the freezing and boiling points of water with 180 intervals between them?

The Fahrenheit scale uses \( 32^\circ \text{F} \) for freezing and \( 212^\circ \text{F} \) for boiling of water, with 180 intervals (\( 212 - 32 = 180 \)) between these points.

Kelvin
Celsius
Fahrenheit
Centigrade
3

Which material is preferred as a coolant in radiators due to its high capacity to absorb heat?

Water is preferred as a coolant in radiators because of its high specific heat capacity, allowing it to absorb significant heat with minimal temperature rise.

Alcohol
Water
Mercury
Oil
2

A \( 0.45 \, \text{kg} \) copper block at \( 200^\circ \text{C} \) is placed in \( 1.1 \, \text{kg} \) water at \( 23^\circ \text{C} \) in a \( 0.2 \, \text{kg} \) brass calorimeter at \( 23^\circ \text{C} \). What is the final temperature? (Specific heat of copper = \( 386 \, \text{J kg}^{-1} \text{K}^{-1} \), water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), brass = \( 386 \, \text{J kg}^{-1} \text{K}^{-1} \))

Heat lost = Heat gained.

\( 0.45 \times 386 \times (200 - T) = (1.1 \times 4186 + 0.2 \times 386) \times (T - 23) \).

\( 34740 - 173.7 T = (4604.6 + 77.2) \times (T - 23) = 4681.8 T - 107678.4 \).

\( 34740 + 107678.4 = 4681.8 T + 173.7 T \).

\( 142418.4 = 4855.5 T \Rightarrow T \approx 29.33^\circ \text{C} \approx 29.3^\circ \text{C} \).

29.3°C
29°C
30°C
28°C
1

What is the primary factor affecting the volume expansion of a gas compared to solids and liquids?

Gases expand much more than solids and liquids due to their higher coefficient of volume expansion, which is temperature-dependent and significantly larger (Section 10.5).

Density
Specific heat
Pressure
Coefficient of volume expansion
4

Which property of a material explains why sand heats up faster than water under the same sunlight?

Specific heat capacity determines how much heat is needed to raise the temperature of a material. Sand has a lower specific heat capacity than water, so it heats up faster.

Specific heat capacity
Latent heat
Density
Thermal expansion
1

How much heat is required to raise \( 0.5 \, \text{kg} \) of water from \( 20^\circ \text{C} \) to \( 80^\circ \text{C} \) and then convert \( 0.2 \, \text{kg} \) of it to steam at \( 100^\circ \text{C} \)? (Specific heat of water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), latent heat of vaporization = \( 2.256 \times 10^6 \, \text{J kg}^{-1} \))

\( Q_1 = 0.5 \times 4186 \times (80 - 20) = 0.5 \times 4186 \times 60 = 125580 \, \text{J} \) (to 80°C).

\( Q_2 = 0.5 \times 4186 \times (100 - 80) = 0.5 \times 4186 \times 20 = 41860 \, \text{J} \) (to 100°C).

\( Q_3 = 0.2 \times 2.256 \times 10^6 = 451200 \, \text{J} \) (vaporization of 0.2 kg).

Total: \( Q = 125580 + 41860 + 451200 = 618640 \, \text{J} = 618.64 \, \text{kJ} \).

615 kJ
618.64 kJ
620 kJ
610 kJ
2

A gas at \( 300 \, \text{K} \) and \( 1 \, \text{atm} \) occupies \( 3 \, \text{L} \). If it is heated to \( 450 \, \text{K} \) while the volume is adjusted to \( 4.5 \, \text{L} \), what is the final pressure?

Given: \( T_1 = 300 \, \text{K} \), \( P_1 = 1 \, \text{atm} \), \( V_1 = 3 \, \text{L} \), \( T_2 = 450 \, \text{K} \), \( V_2 = 4.5 \, \text{L} \).

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

\( P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 1 \times \frac{3}{4.5} \times \frac{450}{300} = 1 \times \frac{2}{3} \times 1.5 = 1 \, \text{atm} \).

0.9 atm
1.1 atm
1 atm
1.2 atm
3

A gas at \( 2.8 \, \text{atm} \) and \( 97^\circ \text{C} \) occupies \( 3.8 \, \text{L} \). If the temperature is raised to \( 147^\circ \text{C} \) and volume reduced to \( 2.5 \, \text{L} \), what is the final pressure?

Given: \( P_1 = 2.8 \, \text{atm} \), \( T_1 = 97^\circ \text{C} = 370 \, \text{K} \), \( V_1 = 3.8 \, \text{L} \), \( T_2 = 147^\circ \text{C} = 420 \, \text{K} \), \( V_2 = 2.5 \, \text{L} \).

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

\( P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 2.8 \times \frac{3.8}{2.5} \times \frac{420}{370} \).

\( P_2 = 2.8 \times 1.52 \times 1.1351 \approx 4.83 \, \text{atm} \).

4.8 atm
4.9 atm
4.83 atm
4.7 atm
3

A steel plate has an area of \( 1.2 \, \text{m}^2 \) at \( 45^\circ \text{C} \). What is the decrease in area when cooled to \( 5^\circ \text{C} \)? (\( \alpha_l = 1.2 \times 10^{-5} \, \text{K}^{-1} \))

Given: \( A_0 = 1.2 \, \text{m}^2 \), \( \Delta T = 5 - 45 = -40^\circ \text{C} \), \( \alpha_l = 1.2 \times 10^{-5} \, \text{K}^{-1} \).

Area expansion: \( \Delta A = A_0 \times 2 \alpha_l \Delta T = 1.2 \times 2 \times 1.2 \times 10^{-5} \times (-40) \).

\( \Delta A = 1.2 \times 2.4 \times 10^{-5} \times (-40) = -0.001152 \, \text{m}^2 \) (decrease of \( 0.001152 \, \text{m}^2 \)).

0.0011 m²
0.0012 m²
0.00115 m²
0.001152 m²
4

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