Question

The temperature of source of a Carnot engine of efficiency $20\%$ when the heat exhausted is at $240 \,K$ is

• A. $1200\, K$
• B. $600\, K$
• C. $540\, K$
• D. $300\, K$

Answer 1

Answer: (D)

$300\, K$

Answer 2

The efficiency of a Carnot engine is
$\eta=-1-\frac{T_{2}}{T_{1}}$
where $T_{1}$ and $T_{2}$ are the temperatures (in kelvin) of the source and the sink respectively
Here, $\eta=20\% =\frac{20}{100}=\frac{1}{5}$,
$T_{1}=?, T_{2}=240\,K$
$\therefore \frac{1}{5}=1-\frac{240}{T_{1}}$
or $\frac{240}{T_{1}}=1-\frac{1}{5}=\frac{4}{5}$
or $T_{1}=\frac{5}{4}\left(240\right)$
$=300\, K$