Heat & Thermodynamics for Competitive Exam Success

Heat & Thermodynamics

Q. What is the change in internal energy of a gas that absorbs 300 J of heat and does 100 J of work on the surroundings? (2020)

A. 200 J
B. 300 J
C. 400 J
D. 500 J

Solution

Change in internal energy (ΔU) = Q - W = 300 J - 100 J = 200 J.

Correct Answer: A — 200 J

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Q. What is the change in internal energy of a system if 200 J of heat is added and 50 J of work is done by the system?

A. 150 J
B. 250 J
C. 200 J
D. 100 J

Solution

Using the first law of thermodynamics, ΔU = Q - W = 200 J - 50 J = 150 J.

Correct Answer: A — 150 J

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Q. What is the change in internal energy of a system if 300 J of heat is added and 100 J of work is done by the system? (2021)

A. 200 J
B. 300 J
C. 400 J
D. 500 J

Solution

Using the first law of thermodynamics: ΔU = Q - W = 300 J - 100 J = 200 J.

Correct Answer: A — 200 J

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Q. What is the efficiency of a Carnot engine operating between 300 K and 600 K? (2022)

A. 0.5
B. 0.33
C. 0.25
D. 0.67

Solution

Efficiency = 1 - (T_c/T_h) = 1 - (300/600) = 0.5.

Correct Answer: A — 0.5

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Q. What is the efficiency of a Carnot engine operating between 500 K and 300 K? (2022)

A. 0.4
B. 0.5
C. 0.6
D. 0.7

Solution

Efficiency = 1 - (T2/T1) = 1 - (300/500) = 0.4 or 40%.

Correct Answer: C — 0.6

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Q. What is the efficiency of a Carnot engine operating between a hot reservoir at 600 K and a cold reservoir at 300 K?

A. 0.5
B. 0.33
C. 0.25
D. 0.75

Solution

Efficiency (η) = 1 - (T_c/T_h) = 1 - (300/600) = 0.5 or 50%.

Correct Answer: B — 0.33

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Q. What is the efficiency of a Carnot engine operating between temperatures of 500 K and 300 K?

A. 0.4
B. 0.5
C. 0.6
D. 0.7

Solution

Efficiency (η) = 1 - (T2/T1) = 1 - (300/500) = 0.4 or 40%.

Correct Answer: C — 0.6

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Q. What is the final temperature when 200 g of ice at 0°C is added to 100 g of water at 80°C? (Specific heat of water = 4.2 J/g°C) (2020)

A. 0°C
B. 20°C
C. 40°C
D. 80°C

Solution

Using the principle of conservation of energy, the heat lost by water equals the heat gained by ice. Solving gives a final temperature of 20°C.

Correct Answer: B — 20°C

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Q. What is the final temperature when 200 g of ice at 0°C is added to 100 g of water at 80°C? (Assume no heat loss to the surroundings) (2023)

A. 0°C
B. 20°C
C. 40°C
D. 80°C

Solution

Using heat balance: m_ice * L_f + m_water * c * (T_final - 80) = 0. Solving gives T_final = 20°C.

Correct Answer: B — 20°C

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Q. What is the heat required to raise the temperature of 2 kg of aluminum from 25°C to 75°C? (Specific heat of aluminum = 0.9 J/g°C) (2022)

A. 9000 J
B. 18000 J
C. 36000 J
D. 4500 J

Solution

Heat required Q = mcΔT = (2000 g)(0.9 J/g°C)(50°C) = 90000 J.

Correct Answer: B — 18000 J

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Q. What is the latent heat of fusion if 2000 J of energy is required to melt 0.5 kg of ice? (2021)

A. 2000 J/kg
B. 4000 J/kg
C. 1000 J/kg
D. 500 J/kg

Solution

Latent heat of fusion (L) = Q / m = 2000 J / 0.5 kg = 4000 J/kg.

Correct Answer: A — 2000 J/kg

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Q. What is the latent heat of fusion if 2000 J of heat is required to melt 500 g of ice? (2022)

A. 2000 J/kg
B. 4000 J/kg
C. 1000 J/kg
D. 3000 J/kg

Solution

Latent heat of fusion (L) = Q/m = 2000 J / 0.5 kg = 4000 J/kg.

Correct Answer: A — 2000 J/kg

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Q. What is the latent heat of fusion of ice if 80 calories are required to melt 20 grams of ice? (2020)

A. 40 cal/g
B. 60 cal/g
C. 80 cal/g
D. 100 cal/g

Solution

Latent heat of fusion (L_f) = Q / m = 80 cal / 20 g = 4 cal/g.

Correct Answer: C — 80 cal/g

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Q. What is the latent heat of fusion of ice if 80 g of ice at 0°C melts to water at 0°C, absorbing 3360 J of heat? (2022)

A. 40 J/g
B. 60 J/g
C. 80 J/g
D. 100 J/g

Solution

Latent heat of fusion (L) = Q/m = 3360 J / 80 g = 42 J/g.

Correct Answer: C — 80 J/g

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Q. What is the latent heat of vaporization if 500 g of water at 100°C is converted to steam at the same temperature, absorbing 2260 kJ of heat? (2023)

A. 2260 J/g
B. 450 J/g
C. 334 J/g
D. 1000 J/g

Solution

Latent heat of vaporization (L) = Q / m = 2260 kJ / 500 g = 4.52 kJ/g = 2260 J/g.

Correct Answer: A — 2260 J/g

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Q. What is the latent heat of vaporization of water if 540 J of energy is required to convert 100 g of water at 100°C to steam? (2021)

A. 540 J/g
B. 54 J/g
C. 5.4 J/g
D. 0.54 J/g

Solution

Latent heat of vaporization (L) = Q/m = 540 J / 100 g = 5.4 J/g.

Correct Answer: A — 540 J/g

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Q. What is the specific heat capacity of a substance if it requires 500 J to raise the temperature of 2 kg of the substance by 5°C? (2021)

A. 20 J/kg°C
B. 50 J/kg°C
C. 100 J/kg°C
D. 200 J/kg°C

Solution

Specific heat capacity (c) = Q / (m * ΔT) = 500 J / (2 kg * 5°C) = 50 J/kg°C.

Correct Answer: B — 50 J/kg°C

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Q. What is the specific heat capacity of a substance if it requires 500 J to raise the temperature of 2 kg of the substance by 10°C? (2021)

A. 25 J/kg°C
B. 50 J/kg°C
C. 100 J/kg°C
D. 200 J/kg°C

Solution

Specific heat capacity (c) = Q / (m * ΔT) = 500 J / (2 kg * 10°C) = 25 J/kg°C.

Correct Answer: B — 50 J/kg°C

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Q. What is the specific heat capacity of a substance if it takes 500 J to raise the temperature of 2 kg of the substance by 5°C? (2021)

A. 20 J/kg°C
B. 50 J/kg°C
C. 100 J/kg°C
D. 200 J/kg°C

Solution

Specific heat capacity (c) = Q / (m * ΔT) = 500 J / (2 kg * 5°C) = 50 J/kg°C

Correct Answer: B — 50 J/kg°C

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Q. What is the specific heat capacity of water if it takes 4200 J to raise the temperature of 1 kg of water by 1°C? (2021)

A. 2100 J/kg°C
B. 4200 J/kg°C
C. 8400 J/kg°C
D. 1000 J/kg°C

Solution

The specific heat capacity of water is defined as the amount of heat required to raise the temperature of 1 kg of water by 1°C, which is given as 4200 J/kg°C.

Correct Answer: B — 4200 J/kg°C

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Q. What is the temperature in Celsius if the temperature in Kelvin is 273 K?

A. 0°C
B. 100°C
C. -273°C
D. 273°C

Solution

To convert Kelvin to Celsius, use the formula: °C = K - 273. Therefore, 273 K - 273 = 0°C.

Correct Answer: A — 0°C

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Q. What is the temperature of a gas in Kelvin if its temperature in Celsius is 25°C? (2019)

A. 273 K
B. 298 K
C. 310 K
D. 325 K

Solution

Temperature in Kelvin (K) = Temperature in Celsius (°C) + 273.15 = 25 + 273.15 = 298.15 K.

Correct Answer: B — 298 K

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Q. What is the thermal conductivity of a material if a 1 m² area of it conducts 100 J of heat in 10 seconds with a temperature difference of 20°C?

A. 2 W/m°C
B. 5 W/m°C
C. 10 W/m°C
D. 20 W/m°C

Solution

Using the formula Q = kA(T1-T2)t, we find k = Q / (A(T1-T2)t) = 100 J / (1 m² * 20°C * 10 s) = 0.5 W/m°C.

Correct Answer: B — 5 W/m°C

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Q. What is the work done by an ideal gas during an isobaric expansion if it expands from 2 L to 5 L at a pressure of 3 atm? (2022)

A. 9 J
B. 12 J
C. 15 J
D. 18 J

Solution

Work done (W) = P * ΔV = 3 atm * (5 L - 2 L) = 3 atm * 3 L = 9 atm·L = 15 J (1 atm·L = 101.325 J)

Correct Answer: C — 15 J

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Q. What is the work done by an ideal gas during an isobaric expansion if the pressure is 5 atm and the volume changes from 2 L to 5 L? (2019)

A. 15 L·atm
B. 10 L·atm
C. 5 L·atm
D. 20 L·atm

Solution

Work done (W) = PΔV = P(V_final - V_initial) = 5 atm * (5 L - 2 L) = 15 L·atm.

Correct Answer: A — 15 L·atm

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Q. What is the work done on a gas when it is compressed from 4 L to 2 L at a constant pressure of 2 atm?

A. 4 L·atm
B. 2 L·atm
C. 8 L·atm
D. 0 L·atm

Solution

Work done (W) = P * ΔV = 2 atm * (2 L - 4 L) = 2 atm * (-2 L) = -4 L·atm (work done on the gas is positive).

Correct Answer: A — 4 L·atm

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Q. What is the work done on a gas when it is compressed from 5 L to 2 L at a constant pressure of 200 kPa? (2022)

A. 600 J
B. 800 J
C. 1000 J
D. 1200 J

Solution

Work done (W) = P * ΔV = 200 kPa * (2 L - 5 L) = 200 kPa * (-3 L) = -600 J (work done on the gas).

Correct Answer: C — 1000 J

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Q. What is the work done when 1 mole of an ideal gas expands isothermally from 10 L to 20 L at 300 K? (2022)

A. 300 J
B. 600 J
C. 150 J
D. 0 J

Solution

Work done W = nRT ln(Vf/Vi) = 1 * 8.314 * 300 * ln(20/10) = 600 J.

Correct Answer: B — 600 J

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Q. What is the work done when 1 mole of an ideal gas expands isothermally from 10 L to 20 L at a temperature of 300 K? (2020)

A. 1.5 kJ
B. 2.5 kJ
C. 3.5 kJ
D. 4.5 kJ

Solution

Work done (W) = nRT ln(Vf/Vi) = 1 mol * 8.314 J/(mol K) * 300 K * ln(20/10) = 2.5 kJ.

Correct Answer: B — 2.5 kJ

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Q. What is the work done when 1 mole of an ideal gas expands isothermally from 10 L to 20 L at 300 K? (R = 8.31 J/mol·K) (2023)

A. 0 J
B. 830 J
C. 1660 J
D. 2490 J

Solution

Work done = nRT ln(Vf/Vi) = 1 mol × 8.31 J/mol·K × 300 K × ln(20/10) = 1660 J.

Correct Answer: C — 1660 J

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Heat & Thermodynamics

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