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Q. A Carnot engine has same efficiency between (i) $100\, K$ and $500\, K$, (ii) $T K$ and $900\, K$. The value of $T$ is

Thermodynamics

Solution:

As, $\eta=1-\frac{T_{2}}{T_{1}}$
$\Rightarrow 1-\frac{100}{500}=1-\frac{T}{900}$
$\therefore \frac{T}{900}=\frac{1}{5}$
Or $T=180\, K$