Question

In a heat engine, the temperature of the source and sink is $500K$ and $375K$. If the engine consumes $25\times {10}^5$ per cycle, the work is done per cycle is:

• A. $6.25\times {10}^{5}J$
• B. $3\times {10}^{5}J$
• C. $2.9\times {10}^{5}J$
• D. $4\times {10}^{4}J$

Answer 1

The correct option is A $6.25\times {10}^{5}J$

Given,

$T_1=500K$

$T_2=375K$

$Q_1=25\times {10}^{5}J$

Efficiency of the engine,

$\eta =1-\frac{T_2}{T_1}$

$\eta =1-\frac{375}{500}$

$\eta =1-0.75$

$\eta =0.25=25$%

We know that, $\eta =\frac{W}{Q_1}$

$W=\eta Q_1$

$W=0.25\times 25\times {10}^5$

$W=6.25\times {10}^{5}J$

The correct option is A.