Thermodynamics - Result Question 45
47. An ideal gas heat engine operates in Carnot cycle between $227^{\circ} C$ and $127^{\circ} C$. It absorbs $6 \times 10^{4}$ cals of heat at higher temperature. Amount of heat converted to work is [2005]
(a) $4.8 \times 10^{4} cals$
(b) $6 \times 10^{4} cals$
(c) $2.4 \times 10^{4}$ cals
(d) $1.2 \times 10^{4} cals$
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Answer:
Correct Answer: 47. (d)
Solution:
- (d) We know that efficiency of carnot engine
$=1-\frac{T_2}{T_1}=1-\frac{400}{500}=\frac{1}{5}$
$[\because T_1=(273+227) K=500 K.$
and $.T_2=(273+127) K=400 K]$
Efficiency of Heat engine $=\frac{\text{ Work output }}{\text{ Heat input }}$
or, $\frac{1}{5}=\frac{\text{ work output }}{6 \times 10^{4}}$
$\Rightarrow$ work output $=1.2 \times 10^{4} cal$
