Question
Calculate the efficiency of Carnot's engine working between steam point and ice point.
✓
Answer
Here, steam point $\text{T}_1=100^\circ\text{C}$
$=100+273=373\text{K}$
and ice point $\text{T}_2=0^\circ\text{C}$
$=0+273=273\text{K}$
$\therefore\eta=1-\frac{\text{T}_2}{\text{T}_1}$
$=1-\frac{273}{373}=\frac{100}{373}$
$\therefore\eta=\frac{100}{373}\times100\%=26.81\%$
$=100+273=373\text{K}$
and ice point $\text{T}_2=0^\circ\text{C}$
$=0+273=273\text{K}$
$\therefore\eta=1-\frac{\text{T}_2}{\text{T}_1}$
$=1-\frac{273}{373}=\frac{100}{373}$
$\therefore\eta=\frac{100}{373}\times100\%=26.81\%$
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