Question

A carnot engine takes 900 kcal of heat from a reservoir at 723°C and exhausts it to a sink at 30°C. The work done by the engine is

• A. 2.73 × 106 Cal
• B. 3.73 × 106 Cal
• C. 6.27 × 105 Cal
• D. 3.73 × 105 Cal

Answer 1

Answer: (C)

6.27 × 105 Cal

Answer 2

Here

$Q_{1} = 900kcal = 900 \times 10^{3}Cal = 9 \times 10^{5}Cal$

$T_{1} = 723ºC = 723 + 273 = 996K$

$T_{2} = 30ºC = 30 + 273 = 303K$

$\because\frac{Q_{1}}{Q_{2}} = \frac{T_{1}}{T_{2}}$

$\therefore Q_{2} = \frac{T_{2}}{T_{1}}Q_{1} = \frac{303}{996} \times 9 \times 10^{5} = 2.73 \times 10^{5}Cal$

Now work done by the engine$W = Q_{1} - Q_{2}$

$= 9 \times 10^{5} - 2.73 \times 10^{5}$

$= (9 - 2.73) \times 10^{5} = 6.27 \times 10^{5}Cal$