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  4. A heat engine rejects 600 cal to the sink at 27° C. Amount of work done by the engine will be (Temperature of source is 227° C J=4.2 J / cal )

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Q. A heat engine rejects $600$ cal to the sink at $27^{\circ} C$. Amount of work done by the engine will be (Temperature of source is $227^{\circ} C$ & $J=4.2\, J / cal$ )

Thermodynamics

Solution:

$1-\frac{T_{2}}{T_{1}}=\frac{W}{Q_{1}}$
$1-\frac{300}{500}=\frac{W}{Q_{1}}$
$\frac{W}{Q_{1}}=\frac{2}{5}$
$Q_{1}=\frac{5 W}{2}$
$\because W=Q_{1}-Q_{2}$
Then $Q_{2}=Q_{1}-W$
$Q_{2}=\frac{5 W}{2}-W=\frac{3 W}{2}$
Then $W=\frac{2 Q_{2}}{3}=\frac{2 \times 600}{3}=400 \,cal $
$=400 \times 4.2 \,J =1680\, J$