Question

If carnot engine takes in 3000 kcal of heat from a reservoir at ${627}^{o}C$and gives it to a sink at ${27}^{o}C$. The work done by the engine is :

• A. $4.2\times {10}^{6}J$
• B. $8.4\times {10}^{6}J$
• C. $16.8\times {10}^{6}J$
• D. Zero

Answer 1

Answer: (B)

$8.4\times {10}^{6}J$

Efficiency of a carnot engine : $\eta$ = $1-\frac{T_C}{T_H}$ = $\frac{W_{done}}{Q_{supplied}}$

Given ,
$Q_{supplied}$ = 3000 kcal
$T_C$ = ${27}^{\circ }C$ = 300 K
$T_H$ = ${627}^{\circ }C$ = 900 K

Solving :
$1-\frac{T_C}{T_H}$ = $\frac{W_{done}}{Q_{supplied}}$
$1-\frac{300}{900}$ = $\frac{W_{done}}{3000}$

$W_{done}$ = $\frac{3000\times 2}{3}$ = 2000 kcal = $8.4\times {10}^6$ J (1 Kcal = 4184 J )