Question

A Carnot's engine whose low temperature reservoir is at 7°C has an efficiency 50%. 

By how many degrees should the temperature of the high temperature reservoir be increased to increase the efficiency to 70%?

Answer 1

The Carnot's engine efficiency is given by

$\eta = \frac{T_{high}-T_{low}}{T_{high}}=1-\frac{T_{low}}{T_{high}}$ , where all the temperatures are in Kelvins.

From this relation we deduce that the initial high temperature was $\frac{1}{1-\eta_0}T_{low}$ and the required high temperature is $\frac{1}{1-\eta_1}T_{low}$ and thus the difference is

$\Delta T = (\frac{1}{1-\eta_1}-\frac{1}{1-\eta_0})T_{low} = (\frac{1}{0.3}-\frac{1}{0.5})\cdot 280 \approx 373K$