Heat Engines & Refrigeration — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Heat Engines & Refrigeration MCQs with step-by-step solutions (34 questions). Part of Thermodynamics. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Heat Engines & Refrigeration · easy · theory
A heat engine is a device that converts:
A. Mechanical work entirely into heat
B. Heat from a cold body into a hot body without work
C. Electrical energy into heat
D. Heat energy into mechanical work, operating in a cycle ✓ Correct
Solution: A heat engine takes in heat from a hot reservoir, converts part of it into work, and rejects the remainder to a cold reservoir, repeating the cycle indefinitely.
Q2 — Heat Engines & Refrigeration · easy · theory
The coefficient of performance of a refrigerator, in terms of the heat $Q_C$ extracted from the cold body and the work $W$ supplied, is:
A. $1 - \dfrac{Q_C}{W}$
B. $\dfrac{Q_C}{W}$ ✓ Correct
C. $\dfrac{Q_C}{Q_H}$
D. $\dfrac{W}{Q_C}$
Solution: The useful effect of a refrigerator is the heat removed from the cold space, and the cost is the work input, so $\beta = \dfrac{Q_C}{W}$.
Q3 — Heat Engines & Refrigeration · medium · numerical
A refrigerator absorbs $2000\text{ J}$ of heat from the cold reservoir per cycle and delivers $2500\text{ J}$ to the room. Its coefficient of performance is:
A. $1.25$
B. $5.0$
C. $4.0$ ✓ Correct
D. $0.8$
Solution: Work input $W = Q_H - Q_C = 2500 - 2000 = 500\text{ J}$. So $\beta = \dfrac{Q_C}{W} = \dfrac{2000}{500} = 4.0$.
Q4 — Heat Engines & Refrigeration · medium · theory
The maximum (Carnot) coefficient of performance of a refrigerator working between $T_C$ and $T_H$ is:
A. $\dfrac{T_H}{T_H - T_C}$
B. $\dfrac{T_C}{T_H - T_C}$ ✓ Correct
C. $\dfrac{T_H - T_C}{T_C}$
D. $1 - \dfrac{T_C}{T_H}$
Solution: Using $\dfrac{Q_C}{Q_H} = \dfrac{T_C}{T_H}$ in $\beta = \dfrac{Q_C}{Q_H - Q_C}$ gives $\beta = \dfrac{T_C}{T_H - T_C}$.
Q5 — Heat Engines & Refrigeration · medium · numerical
A refrigerator transfers heat from a cold reservoir at $-3^\circ\text{C}$ to a room at $27^\circ\text{C}$. Its maximum theoretical coefficient of performance is:
A. $10.0$
B. $0.9$
C. $9.0$ ✓ Correct
D. $8.0$
Solution: In kelvin, $T_C = 270\text{ K}$ and $T_H = 300\text{ K}$. So $\beta = \dfrac{270}{300 - 270} = \dfrac{270}{30} = 9.0$.
Q6 — Heat Engines & Refrigeration · hard · theory
The coefficient of performance $\beta$ of a refrigerator and the efficiency $\eta$ of the corresponding heat engine are related by:
A. $\beta = \dfrac{\eta}{1 - \eta}$
B. $\beta = \dfrac{1 - \eta}{\eta}$ ✓ Correct
C. $\beta = \eta$
D. $\beta = 1 - \eta$
Solution: With $\eta = \dfrac{W}{Q_H}$ and $\beta = \dfrac{Q_C}{W} = \dfrac{Q_H - W}{W} = \dfrac{1}{\eta} - 1 = \dfrac{1 - \eta}{\eta}$.
Q7 — Heat Engines & Refrigeration · medium · theory
Unlike the efficiency of a heat engine, the coefficient of performance of a refrigerator:
A. Is always negative
B. Is always exactly $1$
C. Can never exceed $1$
D. Can be greater than $1$ ✓ Correct
Solution: It is a ratio of heat extracted to work supplied, not a fraction of an input, so values of $3$ to $6$ are typical for domestic refrigerators.
Q8 — Heat Engines & Refrigeration · medium · numerical
A refrigerator with a coefficient of performance of $5$ receives $100\text{ J}$ of work per cycle. The heat extracted from the cold chamber is:
A. $600\text{ J}$
B. $500\text{ J}$ ✓ Correct
C. $100\text{ J}$
D. $20\text{ J}$
Solution: $Q_C = \beta W = 5 \times 100 = 500\text{ J}$, and the heat delivered to the room is $Q_H = 500 + 100 = 600\text{ J}$.
Q9 — Heat Engines & Refrigeration · easy · theory
A refrigerator is able to transfer heat from a cold body to a hot body because:
A. The second law does not apply to it
B. External work is supplied to it ✓ Correct
C. Heat naturally flows from cold to hot
D. Its working substance has negative heat capacity
Solution: The Clausius statement forbids such a transfer as a sole result; it becomes possible only when work is done on the system by the compressor.
Q10 — Heat Engines & Refrigeration · hard · theory
The coefficient of performance of a heat pump, defined as $\dfrac{Q_H}{W}$, is related to that of a refrigerator $\beta$ by:
A. $\dfrac{1}{\beta}$
B. $\beta$
C. $\beta - 1$
D. $\beta + 1$ ✓ Correct
Solution: Since $Q_H = Q_C + W$, dividing by $W$ gives $\dfrac{Q_H}{W} = \dfrac{Q_C}{W} + 1 = \beta + 1$.
Q11 — Heat Engines & Refrigeration · medium · theory
The Kelvin-Planck statement of the second law of thermodynamics asserts that it is impossible to:
A. Convert heat completely into work in a cyclic process with a single reservoir ✓ Correct
B. Build any engine with efficiency above $50\%$
C. Compress a gas adiabatically
D. Transfer heat from a hot body to a cold body
Solution: Some heat must always be rejected to a colder reservoir, which is exactly why no heat engine can reach $100\%$ efficiency.
Q12 — Heat Engines & Refrigeration · medium · theory
The Clausius statement of the second law of thermodynamics asserts that it is impossible for:
A. Heat to flow from a hotter to a colder body
B. Heat to flow from a colder to a hotter body without external work ✓ Correct
C. A gas to expand adiabatically
D. A cyclic process to have zero efficiency
Solution: Spontaneous refrigeration is forbidden. The Clausius and Kelvin-Planck statements can be shown to be logically equivalent.
Q13 — Heat Engines & Refrigeration · easy · theory
The efficiency of any practical heat engine is always:
A. Less than that of a Carnot engine working between the same temperatures ✓ Correct
B. Independent of the sink temperature
C. Exactly equal to $100\%$
D. Greater than that of the corresponding Carnot engine
Solution: Real engines suffer friction, turbulence and finite-rate heat transfer, all of which are irreversible and reduce efficiency below the Carnot limit.
Q14 — Heat Engines & Refrigeration · medium · theory
In a refrigerator, the working substance (refrigerant) absorbs heat when it:
A. Flows through the throttle valve
B. Passes through the compressor
C. Evaporates inside the freezer coils at low pressure ✓ Correct
D. Condenses in the coils at the back at high pressure
Solution: The latent heat needed for evaporation is drawn from the contents of the freezer, cooling them. The refrigerant then releases that heat outside when it condenses.
Q15 — Heat Engines & Refrigeration · medium · numerical
A heat engine rejects $600\text{ J}$ of heat while absorbing $900\text{ J}$ per cycle. Its efficiency is:
A. $66.7\%$
B. $50\%$
C. $33.3\%$ ✓ Correct
D. $25\%$
Solution: $W = 900 - 600 = 300\text{ J}$, so $\eta = \dfrac{300}{900} = \dfrac{1}{3} \approx 33.3\%$.
Q16 — Heat Engines & Refrigeration · medium · theory
A device that would convert heat completely into work with no other effect is called a perpetual motion machine of the second kind. Such a device is impossible because it violates:
A. The first law of thermodynamics
B. The second law of thermodynamics ✓ Correct
C. The zeroth law of thermodynamics
D. The law of conservation of mass
Solution: It conserves energy perfectly, so the first law is satisfied. It is the Kelvin-Planck form of the second law that rules it out.
Q17 — Heat Engines & Refrigeration · medium · theory
The purpose of the sink (cold reservoir) in a heat engine is to:
A. Increase the efficiency above $100\%$
B. Supply the heat that is converted into work
C. Receive the rejected heat so that the working substance can return to its initial state ✓ Correct
D. Do work on the working substance
Solution: Without rejecting heat, the working substance could not be restored to its initial state and the cycle could not repeat. This rejection is the unavoidable cost of cyclic operation.
Q18 — Heat Engines & Refrigeration · medium · numerical
A heat engine absorbs $800\text{ J}$ and rejects $500\text{ J}$ per cycle. Its efficiency is:
A. $62.5\%$
B. $25\%$
C. $60\%$
D. $37.5\%$ ✓ Correct
Solution: $W = 800 - 500 = 300\text{ J}$, so $\eta = \dfrac{300}{800} = 0.375 = 37.5\%$.
Q19 — Heat Engines & Refrigeration · easy · numerical
A refrigerator extracts $1200\text{ J}$ from the cold chamber using $300\text{ J}$ of work per cycle. Its coefficient of performance is:
A. $0.25$
B. $5$
C. $4$ ✓ Correct
D. $3$
Solution: $\beta = \dfrac{Q_C}{W} = \dfrac{1200}{300} = 4$.
Q20 — Heat Engines & Refrigeration · medium · numerical
The maximum coefficient of performance of a refrigerator working between $250\text{ K}$ and $300\text{ K}$ is:
A. $0.2$
B. $6$
C. $1.2$
D. $5$ ✓ Correct
Solution: $\beta = \dfrac{T_C}{T_H - T_C} = \dfrac{250}{300 - 250} = \dfrac{250}{50} = 5$.
Q21 — Heat Engines & Refrigeration · medium · numerical
A refrigerator with a coefficient of performance of $6$ removes $1800\text{ J}$ of heat per cycle. The work supplied per cycle is:
A. $600\text{ J}$
B. $300\text{ J}$ ✓ Correct
C. $10800\text{ J}$
D. $1800\text{ J}$
Solution: $W = \dfrac{Q_C}{\beta} = \dfrac{1800}{6} = 300\text{ J}$.
Q22 — Heat Engines & Refrigeration · easy · numerical
A heat engine absorbs $1000\text{ J}$ and rejects $600\text{ J}$ per cycle. Its efficiency is:
A. $30\%$
B. $40\%$ ✓ Correct
C. $67\%$
D. $60\%$
Solution: $W = 1000 - 600 = 400\text{ J}$, so $\eta = \dfrac{400}{1000} = 40\%$.
Q23 — Heat Engines & Refrigeration · easy · numerical
A heat engine absorbs $1500\text{ J}$ and delivers $450\text{ J}$ of work per cycle. Its efficiency is:
A. $30\%$ ✓ Correct
B. $33\%$
C. $70\%$
D. $45\%$
Solution: $\eta = \dfrac{W}{Q_H} = \dfrac{450}{1500} = 0.30 = 30\%$.
Q24 — Heat Engines & Refrigeration · easy · numerical
A refrigerator of coefficient of performance $3$ is supplied with $200\text{ J}$ of work per cycle. The heat extracted from the cold chamber is:
A. $800\text{ J}$
B. $67\text{ J}$
C. $600\text{ J}$ ✓ Correct
D. $200\text{ J}$
Solution: $Q_C = \beta W = 3 \times 200 = 600\text{ J}$.
Q25 — Heat Engines & Refrigeration · easy · numerical
A refrigerator removes $900\text{ J}$ of heat using $300\text{ J}$ of work per cycle. Its coefficient of performance is:
A. $4$
B. $2$
C. $0.33$
D. $3$ ✓ Correct
Solution: $\beta = \dfrac{Q_C}{W} = \dfrac{900}{300} = 3$.
Q26 — Heat Engines & Refrigeration · medium · numerical
The maximum coefficient of performance of a refrigerator working between $280\text{ K}$ and $300\text{ K}$ is:
A. $15$
B. $14$ ✓ Correct
C. $0.93$
D. $7$
Solution: $\beta = \dfrac{T_C}{T_H - T_C} = \dfrac{280}{300 - 280} = \dfrac{280}{20} = 14$.
Q27 — Heat Engines & Refrigeration · medium · numerical
The maximum coefficient of performance of a refrigerator working between $200\text{ K}$ and $300\text{ K}$ is:
A. $1.5$
B. $3$
C. $0.5$
D. $2$ ✓ Correct
Solution: $\beta = \dfrac{200}{300 - 200} = \dfrac{200}{100} = 2$.
Q28 — Heat Engines & Refrigeration · hard · numerical
A Carnot engine has an efficiency of $25\%$. The coefficient of performance of the refrigerator working between the same two temperatures is:
A. $0.75$
B. $3$ ✓ Correct
C. $1.33$
D. $4$
Solution: $\beta = \dfrac{1 - \eta}{\eta} = \dfrac{0.75}{0.25} = 3$.
Q29 — Heat Engines & Refrigeration · hard · numerical
A refrigerator has a coefficient of performance of $4$. Used as a heat pump between the same reservoirs, its coefficient of performance would be:
A. $5$ ✓ Correct
B. $4$
C. $0.25$
D. $3$
Solution: Since $Q_H = Q_C + W$, the heat-pump value is $\dfrac{Q_H}{W} = \beta + 1 = 5$.
Q30 — Heat Engines & Refrigeration · medium · numerical
A refrigerator extracts $2400\text{ J}$ from the cold chamber and rejects $3000\text{ J}$ to the room per cycle. Its coefficient of performance is:
A. $0.8$
B. $5$
C. $4$ ✓ Correct
D. $1.25$
Solution: Work input $W = 3000 - 2400 = 600\text{ J}$, so $\beta = \dfrac{2400}{600} = 4$.