Question

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

• A. $6.25\times {10}^5\text{J}$
• B. $100\times {10}^5\text{J}$
• C. $2.5\times {10}^5\text{J}$
• D. $0.625\times {10}^5\text{J}$

Answer 1

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

Temperature of source, $T_1=500\text{K}$
Temperature of sink, $T_2=375\text{K}$
Heat absorbed, $Q_1=25\times {10}^5\text{J}$
Efficiency of a Carnot's heat engine,
$\eta =\frac{W}{Q_1}=1-\frac{T_2}{T_1}$
$\Rightarrow$ Workdone per cycle is,
$W=Q_1(1-\frac{T_2}{T_1})$
$W=25\times {10}^5(1-\frac{375}{500})$
$\therefore W=25\times {10}^5\times 0.25=6.25\times {10}^5\text{J}$