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An ideal gas heat engine operates in a Carnot cycle between $227^\circ C$ and $127^\circ C$. It absorbs 6 kcal at the higher temperature. The amount of heat (in kcal) converted into work is equal to -
A
4.8
B
3.5
C
1.6
D
1.2
Detailed Solution
Convert temperatures to kelvin: $T_1 = 227 + 273 = 500$ K, $T_2 = 127 + 273 = 400$ K.
Efficiency of a Carnot engine: $\eta = 1 - \dfrac{T_2}{T_1} = \dfrac{W}{Q}$
$\eta = 1 - \dfrac{400}{500} = 0.2$
$W = \eta Q = 0.2 \times 6 = 1.2$ kcal
So 1.2 kcal of heat is converted into work.
Efficiency of a Carnot engine: $\eta = 1 - \dfrac{T_2}{T_1} = \dfrac{W}{Q}$
$\eta = 1 - \dfrac{400}{500} = 0.2$
$W = \eta Q = 0.2 \times 6 = 1.2$ kcal
So 1.2 kcal of heat is converted into work.