A Carnot engine whose low-temperature reservoir is at$\mathbf{17}\mathbf{°}\mathbf{C}$has an efficiency of 40%. By how much should the temperature of the high-temperature reservoir be increased to increase the efficiency to 50%?

The temperature of the low-temperature reservoir${\mathrm{T}}_{\mathrm{L}}=17°\mathrm{C}=290\mathrm{K}$

Initial efficiency,$\epsilon =40\%$or$0.40$

Efficiency after temperature increase,$\epsilon =50\%$or$0.50$

We use the concept of efficiency of Carnot’s heat engine to find the temperature increase of the high-temperature reservoir.

Formula:

The efficiency of Carnot’s engine,

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

Where${\mathrm{T}}_{\mathrm{L}}$ is the temperature of the low-temperature reservoir, ${\mathrm{T}}_{\mathrm{H}}$is the temperature of the high-temperature reservoir, and $\mathrm{\epsilon }$is the efficiency of the system

Before temperature increases, the efficiency of Carnot’s engine is$\mathrm{\epsilon }=0.40$

${\mathrm{T}}_{\mathrm{L}}=17°\mathrm{C}=290\mathrm{K}$

Then solving for${\mathrm{T}}_{\mathrm{H}}$using equation (1), we get that

$$\begin{gathered} {\mathrm{T}}_{\mathrm{H}}=\frac{{\mathrm{T}}_{\mathrm{L}}}{1-\mathrm{\epsilon }} \\ =\frac{290\mathrm{K}}{1-0.40} \\ =483\mathrm{K} \end{gathered}$$

If we replace$\mathrm{\epsilon }=0.50$we obtain the final high temperature using the same equation (1) as given:

$$\begin{gathered} {\mathrm{T}}_{\mathrm{H}}\text{'}=\frac{290\mathrm{K}}{1-0.50} \\ =580\mathrm{K} \end{gathered}$$

The difference between both will give the increased temperature of the reservoir as given:

$$\begin{gathered} {\mathrm{T}}_{\mathrm{H}}\text{'}-{\mathrm{T}}_{\mathrm{H}}=580\mathrm{K}-483\mathrm{K} \\ =97\mathrm{K} \end{gathered}$$

Hence, the value of the increase in temperature of the reservoir is 97 K