Chapter 24, Problem 26A

Calculate the ideal efficiency of a heat engine that takes in energy at 800 K and expels heat to a reservoir at 300 K.

To determine

The ideal efficiency of a heat engine for the given values.

Answer to Problem 26A

The ideal efficiency of a heat engine is $63\%$ .

Explanation of Solution

Given:

The temperature of the hot reservoir

  $T_{\text{hot}}=800\text{ K}$

The temperature of the cold reservoir

  $T_{\text{cold}}=300\text{ K}$

Formula used:

The ideal efficiency of a heat engine is given by the expression,

  $\text{Ideal efficiency}=\frac{T_{\text{hot}}-T_{\text{cold}}}{T_{\text{hot}}}$

Calculation:

Substitute the values of the variables in the formula and calculate the ideal efficiency.

  $\begin{array}{c}\text{Ideal efficiency}=\frac{T_{\text{hot}}-T_{\text{cold}}}{T_{\text{hot}}}\\ =\frac{\left(800\text{ K}\right)-\left(300\text{ K}\right)}{\left(800\text{ K}\right)}\\ =0.625\end{array}$

Express the efficiency of the engine in percentage.

$\begin{array}{c}\text{Ideal efficiency}=0.625\times 100\%\\ =62.5\%\\ \approx 63\%\end{array}$

Conclusion:

Thus, the ideal efficiency of a heat engine is 63%.