A carnot engine operation between temperature T₁ and T₂ has efficiency frac{1}{6} . W

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11.Thermodynamics

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A carnot engine operation between temperature $T_1$ and $T_2$ has efficiency $\frac{1}{6}$. When $T_2$ is lowered by $62\, K$, its efficiency increase to $\frac{1}{3}$. Then $T_1$ and $T_2$ are, respectively

A

$310\, K$ and $248\, K$

B

$372\, K$ and $310\, K$

C

$372\, K$ and $330\, K$

D

$330\, K$ and $268\, K$

Solution

By $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}$

$\frac{1}{6}=1-\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}} \Rightarrow \frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}=\frac{5}{6}$

$\ldots(i)$

and $\frac{1}{3}=1-\frac{\left(\mathrm{T}_{2}-62\right)}{\mathrm{T}_{1}}$ $…(ii)$

By solving eq. $(i)$ $\&$ $(ii),$

$\mathrm{T}_{1}=372 \mathrm{K}$

$\mathrm{T}_{2}=310 \mathrm{K}$

Standard 11

Physics

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