Question

A Carnot engine having efficiency $60%$ receives heat from a source at temperature $600~\text{K$. For the same sink temperature, to increase its efficiency to $80%$, the temperature of the source is}

• A. $300~\text{K}$
• B. $900~\text{K}$
• C. $1200~\text{K}$
• D. $720~\text{K}$

Hint:

Use $\eta = 1 - \dfrac{T_C}{T_H}$ and solve step-by-step to find unknown temperature.

Answer 1

The Correct Option is C

Efficiency of Carnot engine: $\eta = 1 - \dfrac{T_C}{T_H}$
For initial: $0.6 = 1 - \dfrac{T_C}{600} \Rightarrow T_C = 240~\text{K}$
New efficiency: $0.8 = 1 - \dfrac{240}{T_H} \Rightarrow \dfrac{240}{T_H} = 0.2$
$\Rightarrow T_H = \dfrac{240}{0.2} = 1200~\text{K}$