A carnot engine has the same efficiency between $800 K$ to $500 K$ and $x\, K$ to $600 K.$ The value of $x$ is ...... $K$
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(b) In first case, $({\eta _1}) = 1 - \frac{{500}}{{800}} = \frac{3}{8}$
and in second case, $({\eta _2}) = 1 - \frac{{600}}{x}$
Since ${\eta _1} = {\eta _2},$ therefore $\frac{3}{8} = 1 - \frac{{600}}{x}$
or $\frac{{600}}{x} = 1 - \frac{3}{8} = \frac{5}{8}$or$x = \frac{{600 \times 8}}{5} = 960K$
and in second case, $({\eta _2}) = 1 - \frac{{600}}{x}$
Since ${\eta _1} = {\eta _2},$ therefore $\frac{3}{8} = 1 - \frac{{600}}{x}$
or $\frac{{600}}{x} = 1 - \frac{3}{8} = \frac{5}{8}$or$x = \frac{{600 \times 8}}{5} = 960K$
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