Thermodynamic Efficiency & Carnot Cycle — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Thermodynamic Efficiency & Carnot Cycle 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 — Thermodynamic Efficiency & Carnot Cycle · easy · theory
The efficiency of a Carnot engine working between source temperature $T_H$ and sink temperature $T_C$ is:
A. $1 - \dfrac{T_H}{T_C}$
B. $1 - \dfrac{T_C}{T_H}$ ✓ Correct
C. $\dfrac{T_H - T_C}{T_C}$
D. $\dfrac{T_C}{T_H}$
Solution: For a Carnot cycle $\dfrac{Q_C}{Q_H} = \dfrac{T_C}{T_H}$, so $\eta = 1 - \dfrac{Q_C}{Q_H} = 1 - \dfrac{T_C}{T_H}$, with temperatures in kelvin.
Q2 — Thermodynamic Efficiency & Carnot Cycle · easy · numerical
The efficiency of a Carnot engine operating between $600\text{ K}$ and $300\text{ K}$ is:
A. $25\%$
B. $50\%$ ✓ Correct
C. $33.3\%$
D. $75\%$
Solution: $\eta = 1 - \dfrac{300}{600} = 1 - 0.5 = 0.50 = 50\%$.
Q3 — Thermodynamic Efficiency & Carnot Cycle · easy · numerical
An ideal Carnot heat engine operates between $500\text{ K}$ and $300\text{ K}$. Its efficiency is:
A. $40\%$ ✓ Correct
B. $50\%$
C. $60\%$
D. $25\%$
Solution: $\eta = 1 - \dfrac{300}{500} = 1 - 0.6 = 0.40 = 40\%$.
Q4 — Thermodynamic Efficiency & Carnot Cycle · easy · theory
The Carnot cycle consists of:
A. Two isobaric and two isochoric processes
B. Two isothermal and two adiabatic processes ✓ Correct
C. Two isothermal and two isochoric processes
D. Four adiabatic processes
Solution: Heat is taken in during isothermal expansion at $T_H$ and rejected during isothermal compression at $T_C$, with two adiabatic steps connecting the two isotherms.
Q5 — Thermodynamic Efficiency & Carnot Cycle · medium · theory
No heat engine working between two given temperatures can be more efficient than a Carnot engine because the Carnot cycle is:
A. Made of a special working substance
B. Operated at very high pressure
C. Completely reversible ✓ Correct
D. Run at very high speed
Solution: Carnot's theorem states that a reversible engine has the maximum possible efficiency between two reservoirs; any irreversibility (friction, finite temperature differences) only lowers it.
Q6 — Thermodynamic Efficiency & Carnot Cycle · medium · theory
The efficiency of a Carnot engine would be $100\%$ only if:
A. The source and sink were at the same temperature
B. The sink were at absolute zero ✓ Correct
C. The source were at absolute zero
D. The working substance were an ideal gas
Solution: $\eta = 1 - \dfrac{T_C}{T_H}$ reaches unity only for $T_C = 0\text{ K}$, which the third law shows is unattainable.
Q7 — Thermodynamic Efficiency & Carnot Cycle · easy · theory
The efficiency of a heat engine in terms of the heat absorbed $Q_H$ and the work done $W$ is:
A. $\dfrac{Q_H}{W}$
B. $\dfrac{Q_H - W}{Q_H}$
C. $\dfrac{W}{Q_C}$
D. $\dfrac{W}{Q_H}$ ✓ Correct
Solution: Efficiency is the useful output divided by the input: $\eta = \dfrac{W}{Q_H} = 1 - \dfrac{Q_C}{Q_H}$.
Q8 — Thermodynamic Efficiency & Carnot Cycle · medium · numerical
A heat engine absorbs $1000\text{ J}$ per cycle and performs $300\text{ J}$ of work. Its efficiency and the heat rejected are respectively:
A. $30\%$ and $1300\text{ J}$
B. $33\%$ and $300\text{ J}$
C. $70\%$ and $700\text{ J}$
D. $30\%$ and $700\text{ J}$ ✓ Correct
Solution: $\eta = \dfrac{300}{1000} = 30\%$. By energy conservation the heat rejected is $Q_C = Q_H - W = 1000 - 300 = 700\text{ J}$.
Q9 — Thermodynamic Efficiency & Carnot Cycle · medium · theory
The efficiency of a Carnot engine depends on:
A. Only the source and sink temperatures ✓ Correct
B. The mass of the working substance
C. The pressure of the working substance
D. The nature of the working substance
Solution: Carnot efficiency is $1 - \dfrac{T_C}{T_H}$, a result that involves no property of the working fluid whatsoever.
Q10 — Thermodynamic Efficiency & Carnot Cycle · easy · numerical
A Carnot engine works between a source at $400\text{ K}$ and a sink at $300\text{ K}$. Its efficiency is:
A. $25\%$ ✓ Correct
B. $50\%$
C. $75\%$
D. $33\%$
Solution: $\eta = 1 - \dfrac{300}{400} = 1 - 0.75 = 0.25 = 25\%$.
Q11 — Thermodynamic Efficiency & Carnot Cycle · medium · theory
For a Carnot engine, the ratio of the heat rejected to the heat absorbed equals:
A. $\dfrac{T_C}{T_H}$ ✓ Correct
B. $\left(\dfrac{T_C}{T_H}\right)^2$
C. $\dfrac{T_H}{T_C}$
D. $1 - \dfrac{T_C}{T_H}$
Solution: This is the defining property of the Carnot cycle: $\dfrac{Q_C}{Q_H} = \dfrac{T_C}{T_H}$, which is what makes the thermodynamic temperature scale possible.
Q12 — Thermodynamic Efficiency & Carnot Cycle · hard · theory
To raise the efficiency of a Carnot engine, it is more effective to:
A. Increase the mass of the working substance
B. Raise the source temperature by the same amount
C. Increase the number of cycles per second
D. Lower the sink temperature by a given amount ✓ Correct
Solution: Differentiating $\eta = 1 - \dfrac{T_C}{T_H}$ shows the gain from lowering $T_C$ by $\Delta T$ exceeds that from raising $T_H$ by the same $\Delta T$, since $T_H > T_C$.
Q13 — Thermodynamic Efficiency & Carnot Cycle · medium · theory
A reversible thermodynamic process is one that:
A. Occurs very rapidly
B. Occurs only at constant temperature
C. Can be made to retrace its path, leaving no change in the system or surroundings ✓ Correct
D. Always involves friction
Solution: A reversible process passes through a continuous succession of equilibrium states with no dissipative effects, so both system and surroundings can be restored exactly.
Q14 — Thermodynamic Efficiency & Carnot Cycle · medium · theory
Which of the following is an irreversible process?
A. An infinitely slow isothermal expansion
B. An infinitely slow adiabatic compression
C. Free expansion of a gas into a vacuum ✓ Correct
D. An ideal Carnot cycle
Solution: Free expansion happens through non-equilibrium states and cannot be reversed without leaving a change in the surroundings, so it is irreversible.
Q15 — Thermodynamic Efficiency & Carnot Cycle · medium · theory
In a Carnot cycle, heat is absorbed by the working substance during the:
A. Isothermal compression at the sink temperature
B. Adiabatic compression
C. Adiabatic expansion
D. Isothermal expansion at the source temperature ✓ Correct
Solution: The adiabatic steps exchange no heat by definition, and the isothermal compression rejects heat to the sink. Only the isothermal expansion at $T_H$ absorbs heat.
Q16 — Thermodynamic Efficiency & Carnot Cycle · medium · numerical
A Carnot engine has an efficiency of $40\%$ with a sink at $300\text{ K}$. The source temperature is:
A. $420\text{ K}$
B. $500\text{ K}$ ✓ Correct
C. $750\text{ K}$
D. $400\text{ K}$
Solution: $0.40 = 1 - \dfrac{300}{T_H} \Rightarrow \dfrac{300}{T_H} = 0.60 \Rightarrow T_H = \dfrac{300}{0.6} = 500\text{ K}$.
Q17 — Thermodynamic Efficiency & Carnot Cycle · medium · theory
The area enclosed by the Carnot cycle on a $P$-$V$ diagram represents:
A. The efficiency of the engine
B. The change in internal energy per cycle
C. The heat absorbed from the source only
D. The net work done by the engine per cycle ✓ Correct
Solution: The enclosed area is the net work $W = Q_H - Q_C$ delivered by the engine in one complete cycle.
Q18 — Thermodynamic Efficiency & Carnot Cycle · easy · numerical
A Carnot engine works between $800\text{ K}$ and $400\text{ K}$. Its efficiency is:
A. $75\%$
B. $50\%$ ✓ Correct
C. $100\%$
D. $25\%$
Solution: $\eta = 1 - \dfrac{400}{800} = 1 - 0.5 = 50\%$.
Q19 — Thermodynamic Efficiency & Carnot Cycle · medium · numerical
A Carnot engine has an efficiency of $60\%$ with a sink at $300\text{ K}$. The source temperature is:
A. $750\text{ K}$ ✓ Correct
B. $500\text{ K}$
C. $180\text{ K}$
D. $480\text{ K}$
Solution: $0.60 = 1 - \dfrac{300}{T_H} \Rightarrow \dfrac{300}{T_H} = 0.40 \Rightarrow T_H = 750\text{ K}$.
Q20 — Thermodynamic Efficiency & Carnot Cycle · medium · numerical
A Carnot engine working between $500\text{ K}$ and $400\text{ K}$ absorbs $1000\text{ J}$ per cycle. The work it delivers is:
A. $200\text{ J}$ ✓ Correct
B. $800\text{ J}$
C. $500\text{ J}$
D. $100\text{ J}$
Solution: $\eta = 1 - \dfrac{400}{500} = 0.20$, so $W = \eta Q_H = 0.20 \times 1000 = 200\text{ J}$.
Q21 — Thermodynamic Efficiency & Carnot Cycle · hard · numerical
A Carnot engine operating between $400\text{ K}$ and $300\text{ K}$ delivers $100\text{ J}$ of work per cycle. The heat it absorbs from the source is:
A. $133\text{ J}$
B. $300\text{ J}$
C. $500\text{ J}$
D. $400\text{ J}$ ✓ Correct
Solution: $\eta = 1 - \dfrac{300}{400} = 0.25$, so $Q_H = \dfrac{W}{\eta} = \dfrac{100}{0.25} = 400\text{ J}$.
Q22 — Thermodynamic Efficiency & Carnot Cycle · easy · numerical
A Carnot engine works between $900\text{ K}$ and $300\text{ K}$. Its efficiency is:
A. $75\%$
B. $50\%$
C. $33.3\%$
D. $66.7\%$ ✓ Correct
Solution: $\eta = 1 - \dfrac{300}{900} = 1 - 0.333 = 0.667 = 66.7\%$.
Q23 — Thermodynamic Efficiency & Carnot Cycle · easy · numerical
A Carnot engine works between $1000\text{ K}$ and $400\text{ K}$. Its efficiency is:
A. $60\%$ ✓ Correct
B. $40\%$
C. $25\%$
D. $50\%$
Solution: $\eta = 1 - \dfrac{400}{1000} = 0.60 = 60\%$.
Q24 — Thermodynamic Efficiency & Carnot Cycle · medium · numerical
A Carnot engine has an efficiency of $25\%$ with a sink at $300\text{ K}$. The source temperature is:
A. $375\text{ K}$
B. $1200\text{ K}$
C. $500\text{ K}$
D. $400\text{ K}$ ✓ Correct
Solution: $0.25 = 1 - \dfrac{300}{T_H} \Rightarrow \dfrac{300}{T_H} = 0.75 \Rightarrow T_H = 400\text{ K}$.
Q25 — Thermodynamic Efficiency & Carnot Cycle · medium · numerical
A Carnot engine of efficiency $50\%$ has a source at $800\text{ K}$. Its sink temperature is:
A. $1600\text{ K}$
B. $600\text{ K}$
C. $400\text{ K}$ ✓ Correct
D. $200\text{ K}$
Solution: $0.50 = 1 - \dfrac{T_C}{800} \Rightarrow T_C = 0.5 \times 800 = 400\text{ K}$.
Q26 — Thermodynamic Efficiency & Carnot Cycle · medium · numerical
A Carnot engine working between $600\text{ K}$ and $300\text{ K}$ absorbs $600\text{ J}$ per cycle. The work delivered is:
A. $200\text{ J}$
B. $150\text{ J}$
C. $300\text{ J}$ ✓ Correct
D. $450\text{ J}$
Solution: $\eta = 1 - \dfrac{300}{600} = 0.5$, so $W = 0.5 \times 600 = 300\text{ J}$.
Q27 — Thermodynamic Efficiency & Carnot Cycle · medium · numerical
A Carnot engine between $500\text{ K}$ and $250\text{ K}$ delivers $250\text{ J}$ of work per cycle. The heat absorbed from the source is:
A. $750\text{ J}$
B. $1000\text{ J}$
C. $500\text{ J}$ ✓ Correct
D. $250\text{ J}$
Solution: $\eta = 1 - \dfrac{250}{500} = 0.5$, so $Q_H = \dfrac{W}{\eta} = \dfrac{250}{0.5} = 500\text{ J}$.
Q28 — Thermodynamic Efficiency & Carnot Cycle · hard · numerical
A Carnot engine between $800\text{ K}$ and $200\text{ K}$ absorbs $1200\text{ J}$ per cycle. The heat rejected to the sink is:
A. $300\text{ J}$ ✓ Correct
B. $600\text{ J}$
C. $900\text{ J}$
D. $400\text{ J}$
Solution: For a Carnot cycle $\dfrac{Q_C}{Q_H} = \dfrac{T_C}{T_H}$, so $Q_C = 1200 \times \dfrac{200}{800} = 300\text{ J}$.
Q29 — Thermodynamic Efficiency & Carnot Cycle · easy · numerical
A Carnot engine operates between $400\text{ K}$ and $200\text{ K}$. Its efficiency is:
A. $100\%$
B. $50\%$ ✓ Correct
C. $25\%$
D. $75\%$
Solution: $\eta = 1 - \dfrac{200}{400} = 0.50 = 50\%$.
Q30 — Thermodynamic Efficiency & Carnot Cycle · medium · numerical
A Carnot engine of efficiency $40\%$ has a source at $500\text{ K}$. Its sink temperature is:
A. $300\text{ K}$ ✓ Correct
B. $350\text{ K}$
C. $200\text{ K}$
D. $250\text{ K}$
Solution: $0.40 = 1 - \dfrac{T_C}{500} \Rightarrow T_C = 0.6 \times 500 = 300\text{ K}$.