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Heat Engines and Carnot Efficiency

24 min Thermodynamics Math

Every engine that converts heat to work is constrained by a simple ratio — the Carnot efficiency.

First Law of Thermodynamics

$$\Delta U = Q - W$$

Change in internal energy = heat added minus work done by the system.

Thermal Efficiency

$$\eta = \frac{W_{\text{out}}}{Q_{\text{in}}} = 1 - \frac{Q_{\text{out}}}{Q_{\text{in}}}$$

Carnot Efficiency (Maximum possible)

$$\eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H}$$

Temperatures in Kelvin. No real engine can exceed this.

Example 1

A power plant: $T_H = 800$ K, $T_C = 300$ K. Max efficiency?

$\eta = 1 - 300/800 = 1 - 0.375 = 62.5\%$.

Example 2

An engine absorbs 1000 J and exhausts 600 J. Find efficiency and work output.

$W = Q_{in} - Q_{out} = 400$ J.

$\eta = 400/1000 = 40\%$.

Example 3

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

1. Carnot efficiency: $T_H = 500$ K, $T_C = 300$ K.

2. Engine absorbs 2000 J, outputs 500 J work. Efficiency?

3. How much heat is exhausted in Problem 2?

4. Can any engine have 100% efficiency? Why?

5. Find $T_C$ if $T_H = 600$ K and Carnot efficiency is 50%.

6. A Carnot engine has $\eta = 0.4$ and absorbs 5000 J. Work done?

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