Heat & Thermodynamics for Competitive Exam Success

Heat & Thermodynamics

Q. A 1 kg block of ice at 0°C is placed in 2 kg of water at 20°C. What will be the final temperature of the mixture? (Specific heat of water = 4.2 kJ/kg°C, Latent heat of fusion of ice = 334 kJ/kg) (2021)

A. 0°C
B. 10°C
C. 20°C
D. 15°C

Solution

Heat lost by water = Heat gained by ice. Calculate to find the final temperature.

Correct Answer: B — 10°C

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Q. A 1 kg block of ice at 0°C is placed in 2 kg of water at 20°C. What will be the final temperature of the mixture? (Assume no heat loss to the surroundings)

A. 0°C
B. 10°C
C. 15°C
D. 20°C

Solution

Using the principle of conservation of energy, the heat lost by water equals the heat gained by ice. Final temperature can be calculated to be approximately 10°C.

Correct Answer: B — 10°C

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Q. A 1 kg block of ice at 0°C is placed in 2 kg of water at 80°C. What will be the final temperature of the mixture? (Assume no heat loss to the surroundings) (2019)

A. 0°C
B. 40°C
C. 60°C
D. 80°C

Solution

Using the principle of conservation of energy, the heat lost by water equals the heat gained by ice. The final temperature will be 0°C as the ice will melt.

Correct Answer: B — 40°C

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Q. A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. Assuming no heat loss to the surroundings, what is the final temperature of the system? (Specific heat of water = 4.18 kJ/kg°C, specific heat of metal = 0.9 kJ/kg°C) (2020)

A. 25°C
B. 30°C
C. 35°C
D. 40°C

Solution

Using the principle of conservation of energy, set heat lost by metal equal to heat gained by water to find the final temperature.

Correct Answer: C — 35°C

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Q. A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. Assuming no heat loss to the surroundings, what will be the final temperature? (Specific heat of water = 4.2 kJ/kg°C) (2022)

A. 25°C
B. 30°C
C. 35°C
D. 40°C

Solution

Using the principle of conservation of energy, set heat lost by metal equal to heat gained by water to find the final temperature.

Correct Answer: C — 35°C

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Q. A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. If the final temperature of the system is 30°C, what is the specific heat capacity of the metal? (Specific heat of water = 4.18 J/g°C) (2020)

A. 0.5 J/g°C
B. 1.0 J/g°C
C. 1.5 J/g°C
D. 2.0 J/g°C

Solution

Using the principle of conservation of energy, calculate the specific heat capacity of the metal.

Correct Answer: B — 1.0 J/g°C

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Q. A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. What is the final temperature of the system? (Specific heat of water = 4.18 J/g°C, specific heat of metal = 0.9 J/g°C) (2021)

A. 25°C
B. 30°C
C. 35°C
D. 40°C

Solution

Using conservation of energy: m1*c1*(T_initial - T_final) = m2*c2*(T_final - T_initial). Solving gives T_final = 35°C.

Correct Answer: C — 35°C

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Q. A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. What will be the final temperature of the system assuming no heat loss? (2022)

A. 30°C
B. 40°C
C. 50°C
D. 60°C

Solution

Using the principle of conservation of energy: m1*c1*(T_initial1 - T_final) = m2*c2*(T_final - T_initial2). Solving gives T_final = 50°C.

Correct Answer: C — 50°C

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Q. A 100 g piece of metal at 100°C is placed in 200 g of water at 20°C. What will be the final temperature of the system? (Specific heat of water = 4.2 J/g°C, specific heat of metal = 0.5 J/g°C) (2023)

A. 30°C
B. 40°C
C. 50°C
D. 60°C

Solution

Using the heat transfer equation, we can find the final temperature to be 50°C.

Correct Answer: C — 50°C

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Q. A 1000 W heater is used to heat water. How much heat is supplied in 10 minutes? (2020)

A. 600 kJ
B. 300 kJ
C. 100 kJ
D. 150 kJ

Solution

Heat (Q) = Power (P) * Time (t) = 1000 W * 600 s = 600 kJ.

Correct Answer: A — 600 kJ

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Q. A 2 kg block of ice at 0°C is converted to water at 0°C. How much heat is absorbed if the latent heat of fusion of ice is 334,000 J/kg?

A. 668,000 J
B. 334,000 J
C. 167,000 J
D. 500,000 J

Solution

Heat absorbed = mass * latent heat = 2 kg * 334,000 J/kg = 668,000 J.

Correct Answer: A — 668,000 J

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Q. A 2 kg block of metal at 100°C is placed in 1 kg of water at 20°C. What is the final temperature of the system? (Specific heat of water = 4.2 J/g°C, specific heat of metal = 0.9 J/g°C) (2021)

A. 30°C
B. 40°C
C. 50°C
D. 60°C

Solution

Using the heat transfer equation, the final temperature can be calculated to be 50°C.

Correct Answer: C — 50°C

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Q. A 2 kg object is heated from 20°C to 80°C. If the specific heat capacity is 0.5 kJ/kg°C, how much heat is absorbed? (2023)

A. 30 kJ
B. 40 kJ
C. 20 kJ
D. 10 kJ

Solution

Q = mcΔT = 2 kg * 0.5 kJ/kg°C * (80°C - 20°C) = 60 kJ.

Correct Answer: B — 40 kJ

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Q. A 200 g piece of metal at 100°C is placed in 300 g of water at 20°C. What is the final temperature of the system? (Specific heat of water = 4.18 J/g°C, specific heat of metal = 0.9 J/g°C) (2022)

A. 22.5°C
B. 30°C
C. 35°C
D. 40°C

Solution

Using the principle of conservation of energy: m1c1(T_initial - T_final) = m2c2(T_final - T_initial). Solving gives T_final = 35°C.

Correct Answer: C — 35°C

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Q. A 5 kg block of ice at 0°C is converted to water at 0°C. If the latent heat of fusion of ice is 334 kJ/kg, how much heat is absorbed?

A. 1670 kJ
B. 334 kJ
C. 167 kJ
D. 3340 kJ

Solution

Heat absorbed (Q) = m * Lf = 5 kg * 334 kJ/kg = 1670 kJ.

Correct Answer: B — 334 kJ

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Q. A 5 kg block of ice at 0°C is placed in a calorimeter containing 2 kg of water at 80°C. What is the final temperature of the system when thermal equilibrium is reached? (Latent heat of fusion of ice = 334 kJ/kg)

A. 0°C
B. 20°C
C. 40°C
D. 60°C

Solution

The heat lost by water = heat gained by ice. Solving gives the final temperature as 0°C.

Correct Answer: B — 20°C

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Q. A gas expands from 2 L to 5 L at a constant pressure of 1 atm. How much work is done by the gas? (2023)

A. 3 L·atm
B. 5 L·atm
C. 2 L·atm
D. 1 L·atm

Solution

Work done by the gas during expansion at constant pressure is W = PΔV. Here, ΔV = 5 L - 2 L = 3 L, so W = 1 atm * 3 L = 3 L·atm.

Correct Answer: A — 3 L·atm

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Q. A gas expands from a volume of 2 m³ to 5 m³ at a constant pressure of 100 kPa. What is the work done by the gas? (2022)

A. 300 kJ
B. 500 kJ
C. 700 kJ
D. 3000 kJ

Solution

Work done (W) = P * ΔV = 100 kPa * (5 m³ - 2 m³) = 100 kPa * 3 m³ = 300 kJ.

Correct Answer: A — 300 kJ

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Q. A gas expands from volume V1 to V2 at constant pressure P. What is the work done by the gas? (2022)

A. P(V2 - V1)
B. P(V1 + V2)
C. P(V1 * V2)
D. P(V1 - V2)

Solution

Work done (W) = P * ΔV = P * (V2 - V1).

Correct Answer: A — P(V2 - V1)

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Q. A gas expands from volume V1 to V2 at constant temperature. If the initial pressure is P1, what is the final pressure P2? (2020)

A. P1(V1/V2)
B. P1(V2/V1)
C. P1
D. P1/2

Solution

According to Boyle's Law, P1V1 = P2V2, thus P2 = P1(V1/V2).

Correct Answer: A — P1(V1/V2)

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Q. A gas expands isothermally at 300 K and absorbs 600 J of heat. What is the work done by the gas? (2023)

A. 600 J
B. 300 J
C. 900 J
D. 0 J

Solution

In an isothermal process, the work done by the gas is equal to the heat absorbed. Therefore, work done = 600 J.

Correct Answer: A — 600 J

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Q. A gas expands isothermally at 300 K from a volume of 1 m³ to 2 m³. If the pressure at the initial state is 100 kPa, what is the work done by the gas?

A. 0 kJ
B. 100 kJ
C. 150 kJ
D. 200 kJ

Solution

Work done (W) = nRT ln(Vf/Vi). Here, nRT = PVi = 100 kPa * 1 m³ = 100 kJ. W = 100 kJ * ln(2) ≈ 69.3 kJ.

Correct Answer: B — 100 kJ

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Q. A gas expands isothermally at 300 K from a volume of 1 m³ to 2 m³. If the pressure of the gas is 100 kPa, what is the work done by the gas? (2020)

A. 0 kJ
B. 10 kJ
C. 20 kJ
D. 30 kJ

Solution

Work done (W) = P * ΔV = 100 kPa * (2 m³ - 1 m³) = 100 kPa * 1 m³ = 100 kJ = 10 kJ.

Correct Answer: B — 10 kJ

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Q. A gas is compressed adiabatically. Which of the following statements is true? (2023)

A. Temperature increases
B. Pressure decreases
C. Volume increases
D. Internal energy decreases

Solution

In an adiabatic compression, the temperature of the gas increases due to work done on it.

Correct Answer: A — Temperature increases

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Q. A gas is compressed isothermally from a volume of 4 L to 1 L at a constant temperature of 300 K. If the initial pressure is 1 atm, what is the final pressure?

A. 4 atm
B. 3 atm
C. 2 atm
D. 1 atm

Solution

Using Boyle's Law, P1V1 = P2V2, we find P2 = P1 * (V1/V2) = 1 atm * (4 L / 1 L) = 4 atm.

Correct Answer: A — 4 atm

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Q. A metal rod is heated at one end. If the temperature at the heated end is 100°C and the other end is at 20°C, what is the temperature gradient along the rod?

A. 80°C/m
B. 20°C/m
C. 10°C/m
D. 5°C/m

Solution

Temperature gradient = (T_hot - T_cold) / Length. Assuming length is 8 m, gradient = (100°C - 20°C) / 8 m = 10°C/m.

Correct Answer: C — 10°C/m

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Q. A metal rod of length 1 m and cross-sectional area 1 cm² is heated at one end. If the temperature difference between the ends is 100°C, what is the rate of heat transfer through the rod? (Thermal conductivity of the metal = 200 W/m°C)

A. 200 W
B. 400 W
C. 600 W
D. 800 W

Solution

Using Fourier's law: Q/t = kA(ΔT/L) = 200 W/m°C * 0.0001 m² * (100°C/1 m) = 200 W.

Correct Answer: A — 200 W

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Q. A refrigerator operates between 250 K and 300 K. What is its coefficient of performance (COP)? (2022)

A. 1.2
B. 2.0
C. 3.0
D. 4.0

Solution

COP = T_cold / (T_hot - T_cold) = 250 K / (300 K - 250 K) = 250 / 50 = 5.0.

Correct Answer: C — 3.0

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Q. A refrigerator removes heat from the inside at a rate of 200 J/s and expels it to the surroundings at a rate of 250 J/s. What is the coefficient of performance (COP) of the refrigerator? (2023)

A. 0.8
B. 1.25
C. 1.5
D. 2.0

Solution

COP = Q_in / W = 200 J/s / (250 J/s - 200 J/s) = 200 / 50 = 4.

Correct Answer: B — 1.25

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Q. A refrigerator removes heat from the inside at a rate of 200 J/s and expels it to the outside at a rate of 250 J/s. What is the coefficient of performance (COP) of the refrigerator? (2022)

A. 0.8
B. 1.25
C. 1.5
D. 2.0

Solution

COP = Q_c / W = 200 J/s / (250 J/s - 200 J/s) = 200 / 50 = 4.

Correct Answer: B — 1.25

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Heat & Thermodynamics

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