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Q. An ideal heat engine exhausting heat at $77^{\circ} C$ is to have $30 \%$ efficiency. It must take heat at

Punjab PMETPunjab PMET 2005Thermodynamics

Solution:

Given : $T_{2}=77+273=350\, K$
Efficiency $\eta=\left(1-\frac{T_{2}}{T_{1}}\right) \times 100$
$30=\left(1-\frac{350}{T_{1}}\right) 100$
$1-\frac{350}{T_{1}} =\frac{3}{10}$
$\frac{350}{T_{1}} =\frac{7}{10}$
$T_{1} =\frac{350 \times 10}{7}=500\, K$
$=500-273=227^{\circ} C$