Question

An ideal heat engine working between temperature \(T_1\) and \(T_2\) has an efficiency \(\eta\), the new efficiency if both the source and sink temperature are doubled, will be

Answer 1

In first case when temperature of sink is \(T_2\) and temperature of source is \(T_1\) then efficiency of the Carnot engine will be,

\(\eta_1=\dfrac{T_1-T_2}{T_1}\)

In second case when temperature of both sink and source is doubled then,

\(\eta_2=\dfrac{2T_1-2T_2}{2T_1}\)

\(\Rightarrow\eta_2=\dfrac{T_1-T_2}{T_1}\)

\(\Rightarrow\eta_2=\eta\)