Question

A Carnot engine absorbs $6\times {10}^5\text{cal}$ at ${227}^{\circ }\text{C}$. Heat rejected to the sink at ${127}^{\circ }\text{C}$ is

• A. $6\times {10}^5\text{cal}$
• B. $4.8\times {10}^5\text{cal}$
• C. $4\times {10}^5\text{cal}$
• D. $5\times {10}^5\text{cal}$

Answer 1

The correct option is B $4.8\times {10}^5\text{cal}$

It is given that

$Q_1=6\times {10}^5\text{cal}$

$T_1={227}^{\circ }\text{C}=227+273=500\text{K}$

$T_2={127}^{\circ }\text{C}=127+273=400\text{K}$


We know that for Carnot cycle,

$\frac{Q_2}{Q_1}=\frac{T_2}{T_1}$

$\Rightarrow Q_2=\frac{T_2}{T_1}{Q}_1=\frac{400}{500}\times 6\times {10}^5$

$\Rightarrow Q_2=4.8\times {10}^5\text{cal}$

Hence, option $\left(b\right)$ is the correct choice.