Question
A carnot engine absorbs 100 calories per cycle from its source at 1600K. Its efficiency is 60%. Find the temperature of the sink and work done per cycle. Given J = 4.2J/ cal.
✓
Answer
$\text{Q}_1=100\text{cal}, \text{T}_1=1600\text{K},\ \eta=60\%=\frac35$
For glass, $\eta=1-\frac{\text{T}_2}{\text{T}_1},$
$\frac35=1-\frac{\text{T}_2}{1600}$
$\text{T}_2=640\text{K}$
As, $\eta=\frac{\text{W}}{\text{Q}_1},\text{W}=\eta\text{Q}_1$
$=\frac35\times100=60\text{cal}.$
$\text{W}=60\times4.2\text{J}$
$=252\text{J}$
For glass, $\eta=1-\frac{\text{T}_2}{\text{T}_1},$
$\frac35=1-\frac{\text{T}_2}{1600}$
$\text{T}_2=640\text{K}$
As, $\eta=\frac{\text{W}}{\text{Q}_1},\text{W}=\eta\text{Q}_1$
$=\frac35\times100=60\text{cal}.$
$\text{W}=60\times4.2\text{J}$
$=252\text{J}$
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