Thermal Properties of Matter Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

How much heat is needed to convert \( 0.2 \, \text{kg} \) of water at \( 100^\circ \text{C} \) to steam at \( 100^\circ \text{C} \)? (Latent heat of vaporization = \( 2.256 \times 10^6 \, \text{J kg}^{-1} \))

Given: \( m = 0.2 \, \text{kg} \), \( L_v = 2.256 \times 10^6 \, \text{J kg}^{-1} \).

\( Q = m L_v = 0.2 \times 2.256 \times 10^6 = 451200 \, \text{J} = 451.2 \, \text{kJ} \).

450 kJ
451.2 kJ
460 kJ
445 kJ
2

What is the relationship between the Celsius and Kelvin scales?

The Kelvin scale is offset from the Celsius scale by \( 273.15 \), so \( T = t_C + 273.15 \) (Section 10.4), where \( T \) is Kelvin and \( t_C \) is Celsius.

They are identical
\( T = t_C + 273.15 \)
\( t_C = T + 273.15 \)
\( T = t_C - 273.15 \)
2

What is the Kelvin temperature equivalent to \( -23^\circ \text{C} \)?

Relation: \( T = t_C + 273.15 \).

\( T = -23 + 273.15 = 250.15 \, \text{K} \approx 250 \, \text{K} \).

248 K
250 K
252 K
255 K
2

What is the key difference between Celsius and Fahrenheit scales at the boiling point of water?

At the boiling point of water under standard pressure, the Celsius scale reads \( 100^\circ \text{C} \), while the Fahrenheit scale reads \( 212^\circ \text{F} \), reflecting different interval sizes and zero points.

They use the same value
Celsius is \( 100^\circ \), Fahrenheit is \( 212^\circ \)
Celsius is \( 212^\circ \), Fahrenheit is \( 100^\circ \)
Both are absolute scales
2

Which of the following is true about latent heat during a phase change?

Latent heat is the heat absorbed or released per unit mass during a phase change without a temperature change (Section 10.8.1, \( Q = m L \)).

It increases the temperature
It depends on the temperature
It is released during melting
It changes the state without changing temperature
4

A \( 0.2 \, \text{kg} \) aluminium block at \( 120^\circ \text{C} \) is placed in \( 0.8 \, \text{kg} \) of water at \( 20^\circ \text{C} \) in a \( 0.1 \, \text{kg} \) copper calorimeter at \( 20^\circ \text{C} \). What is the final temperature? (Specific heat of aluminium = \( 900 \, \text{J kg}^{-1} \text{K}^{-1} \), water = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \), copper = \( 386 \, \text{J kg}^{-1} \text{K}^{-1} \))

Heat lost = Heat gained.

\( 0.2 \times 900 \times (120 - T) = (0.8 \times 4186 + 0.1 \times 386) \times (T - 20) \).

\( 21600 - 180 T = (3348.8 + 38.6) \times (T - 20) = 3387.4 T - 67748 \).

\( 21600 + 67748 = 3387.4 T + 180 T \).

\( 89348 = 3567.4 T \Rightarrow T \approx 25.04^\circ \text{C} \approx 25^\circ \text{C} \).

25°C
26°C
24°C
27°C
1

A silver sphere of radius \( 8 \, \text{cm} \) at \( 70^\circ \text{C} \) is cooled to \( 20^\circ \text{C} \). What is the decrease in its volume? (\( \alpha_l = 1.9 \times 10^{-5} \, \text{K}^{-1} \))

Given: \( r = 8 \, \text{cm} \), \( V_0 = \frac{4}{3} \pi \times 8^3 = \frac{2048}{3} \pi \, \text{cm}^3 \), \( \Delta T = 20 - 70 = -50^\circ \text{C} \), \( \alpha_l = 1.9 \times 10^{-5} \, \text{K}^{-1} \).

\( \alpha_v = 3 \alpha_l = 3 \times 1.9 \times 10^{-5} = 5.7 \times 10^{-5} \, \text{K}^{-1} \).

\( \Delta V = V_0 \alpha_v \Delta T = \frac{2048}{3} \pi \times 5.7 \times 10^{-5} \times (-50) \).

\( \Delta V = \frac{2048}{3} \pi \times (-2.85 \times 10^{-3}) \approx -6.12 \pi \approx -19.22 \, \text{cm}^3 \) (decrease of \( 19.22 \, \text{cm}^3 \)).

19 cm³
19.2 cm³
19.3 cm³
19.22 cm³
4

A steel rod has a length of \( 1 \, \text{m} \) at \( 20^\circ \text{C} \). What will be its length at \( 120^\circ \text{C} \) if the coefficient of linear expansion of steel is \( 1.2 \times 10^{-5} \, \text{K}^{-1} \)?

Given: \( L_0 = 1 \, \text{m} \), \( T_1 = 20^\circ \text{C} \), \( T_2 = 120^\circ \text{C} \), \( \alpha_l = 1.2 \times 10^{-5} \, \text{K}^{-1} \).

\( \Delta T = 120 - 20 = 100^\circ \text{C} \).

Linear expansion: \( \Delta L = L_0 \alpha_l \Delta T = 1 \times 1.2 \times 10^{-5} \times 100 = 1.2 \times 10^{-3} \, \text{m} \).

New length: \( L = L_0 + \Delta L = 1 + 0.0012 = 1.0012 \, \text{m} \).

1.00012 m
1.0012 m
1.012 m
1.002 m
2

How much heat is required to melt \( 0.3 \, \text{kg} \) of ice at \( 0^\circ \text{C} \) to water at \( 0^\circ \text{C} \)? (Latent heat of fusion = \( 3.33 \times 10^5 \, \text{J kg}^{-1} \))

Given: \( m = 0.3 \, \text{kg} \), \( L_f = 3.33 \times 10^5 \, \text{J kg}^{-1} \).

\( Q = m L_f = 0.3 \times 3.33 \times 10^5 = 99900 \, \text{J} = 99.9 \, \text{kJ} \).

95 kJ
99.9 kJ
100 kJ
90 kJ
2

How much heat is required to convert \( 0.5 \, \text{kg} \) of ice at \( -20^\circ \text{C} \) to water at \( 10^\circ \text{C} \)? (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 = \( 4186 \, \text{J kg}^{-1} \text{K}^{-1} \))

\( Q_1 = 0.5 \times 2100 \times 20 = 21000 \, \text{J} \) (ice from -20°C to 0°C).

\( Q_2 = 0.5 \times 3.35 \times 10^5 = 167500 \, \text{J} \) (melting).

\( Q_3 = 0.5 \times 4186 \times 10 = 20930 \, \text{J} \) (water from 0°C to 10°C).

Total: \( Q = 21000 + 167500 + 20930 = 208430 \, \text{J} = 208.43 \, \text{kJ} \).

208 kJ
208.43 kJ
210 kJ
205 kJ
2

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0