Practice Problems
Work through these problems to test your mastery. Undergraduate problems (1–3) build procedural fluency; PhD problems (4–5) require second-law and endoreversible analysis. Reveal solutions only after attempting each problem.
Test your mastery of the Carnot cycle. Problems 1–3 are standard undergraduate level (as in LibreTexts); Problems 4–5 require PhD-level second-law and endoreversible analysis. Expand solutions only after attempting each problem.
Problem 1 — Efficiency and Reservoir Temperature
A Carnot engine currently operates at 40% efficiency, exhausting heat to a cold reservoir at $T_L = 298$ K. To what temperature must the hot reservoir be raised to increase the efficiency to 65%?
Show Solution
From $\eta = 1 - T_L/T_H$ rearrange to $T_H = T_L/(1-\eta)$.
Current: $T_H = 298/0.60 = \mathbf{497\text{ K}}$ (this is the present hot-reservoir temperature).
Required: $T_H = 298/0.35 = \mathbf{851\text{ K}}$ (~578°C).
The hot reservoir must be raised by 354 K to achieve the 25-percentage-point efficiency gain.
Problem 2 — Cold Reservoir Temperature from Heat Data
A Carnot engine absorbs $Q_H = 1.00$ kJ of heat from a hot reservoir at $T_H = 300$ K and exhausts $|Q_L| = 400$ J per cycle. What is the temperature of the cold reservoir?
Show Solution
Thermal efficiency: $\eta = 1 - 400/1000 = 0.60$.
From $\eta = 1 - T_L/T_H$: $T_L = T_H(1-\eta) = 300 \times 0.40 = \mathbf{120\text{ K}}$.
This ${-153}$°C cold reservoir is extremely cold, which is precisely why 60% efficiency is achievable even with a modest $T_H = 300$ K source.
Problem 3 — Carnot Heat Pump Power
An indoor heater operating on the reversed Carnot cycle delivers heat at $\dot{Q}_H = 30$ kJ/s to maintain an indoor temperature of 72°F (295.4 K) while the outdoor temperature is 30°F (272.0 K). What is the minimum power input required?
Show Solution
Carnot heat pump COP:
$$\text{COP}_{\text{HP}} = \frac{T_H}{T_H - T_L} = \frac{295.4}{295.4-272.0} = \frac{295.4}{23.4} \approx 12.6$$Minimum power: $\dot{W} = \dot{Q}_H/\text{COP} = 30/12.6 \approx \mathbf{2.38\text{ kW}}$.
This is only 7.9% of the 30 kW a resistive heater would consume for the same output — the factor of 12.6 amplification is the leverage of the Carnot cycle.
Problem 4 — Second-Law Analysis of a Real Power Plant
A coal-fired steam power plant has $T_H = 873$ K (600°C), $T_L = T_0 = 303$ K (30°C), and an actual thermal efficiency of 38%. Per 1 kJ of heat input $Q_H$: (a) find $\eta_C$ and $\eta_{CA}$; (b) find the second-law efficiency $\eta_{II}$; (c) find entropy generation $S_{\text{gen}}$ and Gouy-Stodola lost work $W_{\text{lost}}$.
Show Solution
(a) $\eta_C = 1 - 303/873 = 65.3\%$ $\eta_{CA} = 1 - \sqrt{303/873} = 41.1\%$
(b) $\eta_{II} = 0.38/0.653 = 58.2\%$ — the plant captures only 58% of its Carnot potential.
(c) Per kJ $Q_H$: heat rejected $|Q_{L,\text{real}}| = 1 - 0.38 = 0.62$ kJ.
$$S_{\text{gen}} = \frac{|Q_L|}{T_L} - \frac{Q_H}{T_H} = \frac{0.62}{303} - \frac{1.00}{873} = 2.047 \times 10^{-3} - 1.146 \times 10^{-3} = 9.0 \times 10^{-4}\text{ kJ/K}$$$$W_{\text{lost}} = T_0 S_{\text{gen}} = 303 \times 9.0 \times 10^{-4} = \mathbf{0.273\text{ kJ}}\text{ per kJ input}$$27.3% of the input is destroyed as lost work, primarily in the boiler (finite-$\Delta T$ heat exchange from combustion gases to steam) and the turbine (blade friction, non-isentropic expansion).
Problem 5 — Endoreversible Engine: Maximum Power Derivation
An endoreversible engine operates between $T_H = 1000$ K and $T_L = 300$ K with equal heat-exchanger conductances $K_H = K_L = K$. Show that the efficiency at maximum power equals the Curzon-Ahlborn value $\eta_{CA} = 1 - \sqrt{T_L/T_H}$, and calculate it numerically.
Show Solution
Let the internal sub-cycle temperatures be $\tau_H$ and $\tau_L$ (with $T_H > \tau_H > \tau_L > T_L$). Heat flows: $\dot{Q}_H = K(T_H - \tau_H)$, $\dot{Q}_L = K(\tau_L - T_L)$. The interior is reversible: $\dot{Q}_H/\tau_H = \dot{Q}_L/\tau_L$.
Power output: $\dot{W} = \dot{Q}_H - \dot{Q}_L = K(T_H - \tau_H) - K(\tau_L - T_L)$. Using the reversibility constraint to eliminate $\tau_L = \tau_H \dot{Q}_L/\dot{Q}_H$ and substituting the heat-exchanger expressions, then differentiating with respect to $\tau_H$ and setting to zero gives the optimal internal temperatures:
$$\tau_H^* = \frac{\sqrt{T_H}(\sqrt{T_H}+\sqrt{T_L})}{2}, \qquad \tau_L^* = \frac{\sqrt{T_L}(\sqrt{T_H}+\sqrt{T_L})}{2}$$The efficiency at this maximum-power point:
$$\eta^* = 1 - \frac{\tau_L^*}{\tau_H^*} = 1 - \frac{\sqrt{T_L}}{\sqrt{T_H}} = 1 - \sqrt{\frac{T_L}{T_H}} = \eta_{CA}$$Numerically: $\eta_{CA} = 1 - \sqrt{300/1000} = 1 - 0.5477 = \mathbf{45.2\%}$, compared to $\eta_C = 70.0\%$. The maximum power is $\dot{W}_{\max} = K(\sqrt{T_H}-\sqrt{T_L})^2/2 = K(31.62-17.32)^2/2 = 102.3K$ watts.