Question

In a hypothetical nuclear fusion reactor, the fuel is deuterium gas at a $\mathbf{7}\mathbf{\times }\mathbf{100}\mathbf{°}\mathbf{C}$temperature of . If this gas could be used to operate a Carnot engine with ${\mathit{T}}_{\mathbf{L}}\mathbf{=}\mathbf{100}\mathbf{°}\mathbf{C}$, what would be the engine’s efficiency? Take both temperatures to be exact and report your answer to seven significant figures.

Short answer

The efficiency of the engine is 99.99995%

Step-by-step solution

The given data

The temperature of the high-temperature reservoir,$T_H=7\times {10}^8\mathrm{K}$

The temperature of the low-temperature reservoir,${\mathrm{T}}_{\mathrm{L}}=100°\mathrm{C}=373\mathrm{K}$

Understanding the concept of the efficiency of the engine

We use the formula of efficiency of Carnot's heat engine to calculate the efficiency of the engine

Formula:

The efficiency of the Carnot’s engine,

$\mathrm{\epsilon }=1-\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}}$ (2)

Calculation of the engine’s efficiency

The efficiency of Carnot’s engine using equation (1) can be given as:

$$\begin{gathered} \mathrm{\epsilon }=\left(1-\frac{373\mathrm{K}}{7\times {10}^8\mathrm{K}}\right)\times 100\% \\ =\left(1-2.3285\times {10}^{-7}\right)\times 100\% \\ =\left(0.9999995\right)\times 100\% \\ =99.99995\% \end{gathered}$$

Hence, the value of the engine’s efficiency is 99.99995%