Question
A carnot engine operating between temperature $T_1$ and $T_2$ has efficiency $\frac{1}{6}$. When $T_2$ is lowered by $62 K$, its efficiency increases to $\frac{1}{3}$. Then find the values of $T_1$ and $T_2$.
✓
Answer
Efficiency
$\eta=1-\frac{T_2}{T_1}$
In the first case:
$1-\frac{ T _2}{ T _1} =\frac{1}{6} \Rightarrow \frac{ T _2}{ T _1}=\frac{5}{6} $
$T _1 =\frac{6}{5} T _2$
In the second case: $1-\frac{\left(T_2-62\right)}{T_1}=\frac{1}{3} ; \frac{T_2-62}{T_1}=\frac{2}{3}$
$ \frac{ T _2-62}{\left(\frac{6}{5}\right) T _2}=\frac{2}{3}$
$T _2-62=\frac{2}{3} \times \frac{6}{5} T _2=\frac{4}{5} T _2 \Rightarrow T _2=310 K$
$T _1=\frac{6}{5} \times 310=372 K$
$T _1=372 K \text { and } T _2=310 K $
$\eta=1-\frac{T_2}{T_1}$
In the first case:
$1-\frac{ T _2}{ T _1} =\frac{1}{6} \Rightarrow \frac{ T _2}{ T _1}=\frac{5}{6} $
$T _1 =\frac{6}{5} T _2$
In the second case: $1-\frac{\left(T_2-62\right)}{T_1}=\frac{1}{3} ; \frac{T_2-62}{T_1}=\frac{2}{3}$
$ \frac{ T _2-62}{\left(\frac{6}{5}\right) T _2}=\frac{2}{3}$
$T _2-62=\frac{2}{3} \times \frac{6}{5} T _2=\frac{4}{5} T _2 \Rightarrow T _2=310 K$
$T _1=\frac{6}{5} \times 310=372 K$
$T _1=372 K \text { and } T _2=310 K $
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