Question

An ideal gas heat engine operates in Carnot cycle between ${272}^{o}C$ and ${127}^{o}C$. It absorbs $6.0\times {10}^{4}cal$ at the higher temperature. The amount of heat converted into work is equal to

• A. $1.2\times {10}^{4}cal$
• B. $1.6\times {10}^{4}cal$
• C. $3.5\times {10}^{4}cal$
• D. $4.8\times {10}^{4}cal$

Answer 1

The correct option is B $1.6\times {10}^{4}cal$

The efficiency of the cycle is given as $\eta =1-\frac{T_2}{T_1}=\frac{W}{Q_1}$. Here $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink. $Q_1$ is the heat taken up from the source and $W$ is the work done in the process.

Thus, we get $\eta =1-\frac{127+273}{272+273}=1-\frac{400}{545}=0.266=\frac{W}{6\times {10}^4}$
Thus, we calculate $W$ as $1.6\times {10}^{4}cal$