Heat engine

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A Carnot’s engine, with its cold body at $17^{0} C$ has 50% efficiency. If the temperature of its hot body is now increased by $145^{0} C$ , the efficiency becomes

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Solution

$\begin{array}{l} η = 1 - \frac{T_{2}}{T_{1}} = 0 .5 \\ ∴ T_{1} = 2 T_{2} = 2 (17 + 273) = 580 K = 307^{0} C \end{array}$
Now the temperature of hot body is increased by $145^{0} C$
$\begin{array}{l} ∴ {T_{1}}^{'} = 307^{0} + 145^{0} = 452^{0} C = 725 K \\ T_{2} = 17 + 273 = 290 K \\ ∴ η' = (1 - \frac{290}{725}) \times 100 = 60 % \end{array}$

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Heat engine

Remember concepts with our Masterclasses.

A Carnot’s engine, with its cold body at 170C has 50% efficiency. If the temperature of its hot body is now increased by1450C , the efficiency becomes

η=1−T2T1=0.5∴T1=2T2=2(17+273)=580K=3070CNow the temperature of hot body is increased by 1450C∴T1'=3070+1450=4520C=725K T2=17+273=290K∴η'=1−290725×100=60%

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A Carnot’s engine, with its cold body at $17^{0} C$ has 50% efficiency. If the temperature of its hot body is now increased by $145^{0} C$ , the efficiency becomes