Question

The efficiency of Carnots heat engine is 0.5 when the temperature of the source is $T_1$ and that of sink is $T_2$. The efficiency of another Carnots heat engine is also 0.5. The temperature of source and sink of the second engine are respectively

• A. ${2T}_1$, ${2T}_2$
• B. ${2T}_1$, $\frac{T_2}{2}$
• C. $T_1+5$, $T_2$-5
• D. $T_1+10$, $T_2$-10

Answer 1

Answer: (A)

${2T}_1$, ${2T}_2$

For carnot engine-

$\eta \left(efficiency\right)=1-\frac{T_L}{T_H}$

$T_H=$ high temperature (sources)

$T_L=$ low temperature(sink)

For same efficiency, ratio of temperature should remain same.

Option $\left(A\right)2T_{1}and2T_2\frac{2T_1}{2T_2}=\frac{T_1}{T_2}$