Q. If 200 J of heat is added to a system and 50 J of work is done by the system, what is the change in internal energy? (2023)
A.
150 J
B.
250 J
C.
200 J
D.
100 J
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Solution
According to the first law of thermodynamics, ΔU = Q - W = 200 J - 50 J = 150 J.
Correct Answer: A — 150 J
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Q. If 2000 J of heat is added to a system and 500 J of work is done by the system, what is the change in internal energy of the system? (2021)
A.
1500 J
B.
2000 J
C.
2500 J
D.
3000 J
Show solution
Solution
Using the first law of thermodynamics: ΔU = Q - W = 2000 J - 500 J = 1500 J.
Correct Answer: A — 1500 J
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Q. If 500 J of heat is added to a system and 200 J of work is done by the system, what is the change in internal energy of the system?
A.
300 J
B.
500 J
C.
700 J
D.
200 J
Show solution
Solution
According to the first law of thermodynamics, ΔU = Q - W. Here, ΔU = 500 J - 200 J = 300 J.
Correct Answer: A — 300 J
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Q. If 500 J of heat is added to a system and it does 200 J of work, what is the change in internal energy? (2022)
A.
300 J
B.
200 J
C.
500 J
D.
700 J
Show solution
Solution
According to the first law of thermodynamics, ΔU = Q - W. Here, ΔU = 500 J - 200 J = 300 J.
Correct Answer: A — 300 J
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Q. If the latent heat of fusion of ice is 334 J/g, how much heat is required to melt 50 g of ice? (2019)
A.
1670 J
B.
3340 J
C.
5000 J
D.
1000 J
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Solution
Heat required = mass × latent heat = 50 g × 334 J/g = 16700 J.
Correct Answer: A — 1670 J
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Q. If the pressure of a gas is increased while keeping the temperature constant, what happens to its volume? (2020)
A.
It increases
B.
It decreases
C.
It remains the same
D.
It becomes zero
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Solution
According to Boyle's Law, P * V = constant at constant temperature, so if P increases, V must decrease.
Correct Answer: B — It decreases
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Q. If the volume of a gas is doubled at constant temperature, what happens to its pressure?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
According to Boyle's Law, at constant temperature, pressure is inversely proportional to volume. Therefore, if the volume is doubled, the pressure is halved.
Correct Answer: B — It halves
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Q. If the volume of a gas is halved at constant temperature, what happens to its pressure? (2023)
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
Show solution
Solution
According to Boyle's Law, if the volume is halved, the pressure doubles at constant temperature.
Correct Answer: A — It doubles
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Q. If the volume of a gas is halved while keeping the temperature constant, what happens to the pressure of the gas?
A.
Halves
B.
Doubles
C.
Remains the same
D.
Increases by 50%
Show solution
Solution
According to Boyle's Law, if the volume is halved, the pressure doubles.
Correct Answer: B — Doubles
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Q. In a Carnot engine operating between 500 K and 300 K, what is the maximum efficiency of the engine? (2021)
A.
40%
B.
60%
C.
50%
D.
20%
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Solution
Efficiency (η) = 1 - (T_cold / T_hot) = 1 - (300 K / 500 K) = 0.4 or 40%.
Correct Answer: B — 60%
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Q. In a Carnot engine operating between temperatures of 500 K and 300 K, what is the maximum efficiency? (2023)
A.
40%
B.
60%
C.
50%
D.
70%
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Solution
Efficiency (η) = 1 - (T_cold / T_hot) = 1 - (300 K / 500 K) = 0.4 or 40%.
Correct Answer: B — 60%
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Q. In a Carnot engine operating between temperatures of 500 K and 300 K, what is the efficiency of the engine? (2023)
A.
40%
B.
50%
C.
60%
D.
70%
Show solution
Solution
Efficiency (η) = 1 - (T_cold / T_hot) = 1 - (300 K / 500 K) = 0.4 or 40%.
Correct Answer: C — 60%
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Q. In a Carnot engine operating between two heat reservoirs at temperatures 500 K and 300 K, what is the maximum efficiency of the engine? (2023)
A.
40%
B.
60%
C.
80%
D.
20%
Show solution
Solution
Efficiency (η) = 1 - (T_cold / T_hot) = 1 - (300 K / 500 K) = 0.4 or 40%.
Correct Answer: B — 60%
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Q. In an adiabatic process, the temperature of an ideal gas decreases. What happens to its pressure?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Depends on volume
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Solution
In an adiabatic process, as the temperature decreases, the pressure also decreases due to the ideal gas law.
Correct Answer: B — Decreases
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Q. In an isochoric process, what remains constant? (2021)
A.
Pressure
B.
Volume
C.
Temperature
D.
Heat
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Solution
In an isochoric process, the volume of the system remains constant.
Correct Answer: B — Volume
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Q. In an isochoric process, which of the following remains constant? (2022)
A.
Pressure
B.
Volume
C.
Temperature
D.
Internal Energy
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Solution
In an isochoric process, the volume remains constant.
Correct Answer: B — Volume
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Q. What is the change in entropy when 1 kg of water at 100°C is converted to steam at 100°C? (Latent heat of vaporization = 2260 kJ/kg) (2021)
A.
2260 J/K
B.
2260 kJ/K
C.
0 J/K
D.
1130 J/K
Show solution
Solution
ΔS = Q/T = 2260 kJ / 373 K = 6.06 kJ/K.
Correct Answer: A — 2260 J/K
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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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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 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
Show solution
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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