Question

1. A Carnot heat engine, whose sink is at 280 K, has an efficiency of 30%. By how much the temperature of the source be increased to have its efficiency equal to 50% keeping sink temperature constant.

Answer 1

$$\begin{gathered} \eta =1-\frac{T_2}{T_1} \\ \frac{T_2}{T_1}=1-\frac{30}{100}=\frac{7}{10} \\ {T}_1=10\frac{T_2}{7}=\frac{10\times 280}{7}=400K \end{gathered}$$
increase in efficiency $$\begin{gathered} \eta \text{'}=50\%of30\%=15 \\ increasedinefficiency=45\% \\ \eta \text{'}=1-\frac{T_2}{T{\text{'}}_1}\frac{T_2}{T{\text{'}}_1}=1-\eta \text{'}=1-\frac{45}{100}=\frac{11}{20}T\text{'} \end{gathered}$$₁=T₂×2011=280×2011=509.1K increaese in temperature of source = 509.1-400=109.1K