Heat Engines and Carnot Efficiency
Heat engines convert thermal energy into mechanical work, and the Carnot cycle defines the theoretical maximum efficiency any engine can achieve between two temperature reservoirs. The Carnot efficiency η = 1 − TC/TH depends only on the absolute temperatures of the hot and cold reservoirs. Real engines — gasoline, diesel, jet turbines — always fall short of this ideal, but understanding the Carnot limit lets you evaluate and compare engine designs quantitatively.
Heat Engines and Carnot Efficiency
Every engine that converts heat to work is constrained by a simple ratio — the Carnot efficiency.
$$\Delta U = Q - W$$
Change in internal energy = heat added minus work done by the system.
$$\eta = \frac{W_{\text{out}}}{Q_{\text{in}}} = 1 - \frac{Q_{\text{out}}}{Q_{\text{in}}}$$
$$\eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H}$$
Temperatures in Kelvin. No real engine can exceed this.
A power plant: $T_H = 800$ K, $T_C = 300$ K. Max efficiency?
- $\eta = 1 - 300/800 = 1 - 0.375 = 62.5\%$.
An engine absorbs 1000 J and exhausts 600 J. Find efficiency and work output.
- $W = Q_{in} - Q_{out} = 400$ J.
- $\eta = 400/1000 = 40\%$.
What $T_H$ is needed for 70% Carnot efficiency if $T_C = 300$ K?
- $0.70 = 1 - 300/T_H$ → $T_H = 300/0.30 = 1000$ K = 727°C.
Practice Problems
Show Answer Key
1. $1 - 300/500 = 40\%$
2. $500/2000 = 25\%$
3. $2000 - 500 = 1500$ J
4. No — would require $T_C = 0$ K, which is unreachable.
5. $T_C = T_H(1-\eta) = 600(0.5) = 300$ K
6. $W = 0.4 \times 5000 = 2000$ J
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