Question

A Carnot engine efficiency 40% and temperature of sink 300 K to increase up to 60% calculate the change in temperature of the source.

Answer 1

Given:

Carnot engine efficiency, ${\eta }_1=40\%$

Temperature of sink, $T_2=300K$

Efficiency increased to ${\eta }_2=60\%$

To find:

Change in temperature of the source, new $T_1=?$

We know efficiency of Carnot engine is,

$\eta =\frac{T_1-T_2}{T_1}⟹\frac{40}{100}=\frac{T_1-300}{T_1}⟹4T_1=10T_1-3000⟹6T_1=3000⟹T_1=500K$

is the source temperature initially.

Now to increase efficiency to 60$\%$, that is

${\eta }_2=\frac{T_1-T_2}{T_1}⟹\frac{60}{100}=\frac{T_1-300}{T_1}⟹6T_1=10T_1-3000⟹4T_1=3000⟹T_1=750K$

is the new source temperature.

Hence the change in temperature of the source is 250K.