Problem Statement
Solve the thermodynamics problem: A Carnot engine operates between $T_1=500\ \text{K}$ (hot) and $T_2=300\ \text{K}$ (cold). Find the efficiency. The Carnot efficiency is the maximum possible efficiency for any heat engine operating between two temperatures: $$\eta = 1 – \frac{T_2}{T_1} = 1 – \frac{300}{500} = 1 – 0.6 = 0.4 = 40\%$$
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
The Carnot engine is the idealized heat engine operating between two reservoirs at temperatures $T_H$ (hot) and $T_C$ (cold). It represents the maximum possible efficiency for any heat engine operating between those two temperatures — a fundamental result of the Second Law of Thermodynamics. No real engine can exceed Carnot efficiency.
- $\eta_{\text{Carnot}} = 1 – T_C/T_H$ — efficiency depends only on absolute temperatures
- $W = Q_H \eta$ — net work output equals heat absorbed times efficiency
- $Q_C = Q_H(1 – \eta)$ — heat rejected to cold reservoir
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\eta = 1 – \frac{T_2}{T_1} = 1 – \frac{300}{500} = 1 – 0.6 = 0.4 = 40\%$$
$$\eta = 1 – \frac{300}{600} = 1 – 0.5 = 0.50 = 50\%$$
$$\boxed{\eta_{\text{Carnot}} = 1 – \dfrac{T_C}{T_H}}$$
Answer
$$\boxed{\eta_{\text{Carnot}} = 1 – \dfrac{T_C}{T_H}}$$
Physical Interpretation
The Carnot efficiency sets an absolute upper bound imposed by thermodynamics — no real engine, however well engineered, can exceed it. Higher hot-reservoir temperature or lower cold-reservoir temperature both increase efficiency.
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