Thermodynamics - Result Question 47
49. An ideal gas heat engine operates in a Carnot cycle between $227^{\circ} C$ and $127^{\circ} C$. It absorbs 6 $kcal$ at the higher temperature. The amount of heat (in kcal) converted into work is equal to
(a) 1.2
(b) 4.8
(c) 3.5
(d) 1.6
[2002]
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Answer:
Correct Answer: 49. (a)
Solution:
- (a) Efficiency $=\frac{T_1-T_2}{T_1}$
$T_1=227+273=500 K$
$T_2=127+273=400 K$
$\eta=\frac{500-400}{500}=\frac{1}{5}$
Hence, output work $=(\eta) \times$ Heat input $=\frac{1}{5} \times 6=1.2 kcal$
Efficiency of heat engine: $n=\frac{\text{ Work done }(W)}{\text{ Heat in put }(Q_1)}$ $=1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1}$
