Let $\eta_{1}$ is the efficiency of an engine at $T_{1}=447^{\circ} \mathrm{C}$ and $\mathrm{T}_{2}=147^{\circ} \mathrm{C}$ while $\eta_{2}$ is the efficiency at $\mathrm{T}_{1}=947^{\circ} \mathrm{C}$ and $\mathrm{T}_{2}=47^{\circ} \mathrm{C}$ The ratio $\frac{\eta_{1}}{\eta_{2}}$ will be :
Solution
<p>${\eta _1} = 1 - {{420} \over {720}} = {{300} \over {720}}$</p>
<p>And ${\eta _2} = 1 - {{320} \over {1220}} = {{900} \over {1220}}$</p>
<p>$$ \Rightarrow {{{\eta _1}} \over {{\eta _2}}} = {{300} \over {720}} \times {{1220} \over {900}}$$</p>
<p>$\simeq 0.56$</p>
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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