Question

# A Carnot engine takes in $$3000\ kcal$$ of heat from a reservoir at $$627^{\circ}C$$ and gives it to a sink at $$27^{\circ}C$$. The work done by the engine is

A
4.2×106J
B
8.4×106J
C
16.8×106J
D
Zero

Solution

## The correct option is B $$8.4\times 10^{6}J$$Given : $$Q_{1} = 3000 \times 10^{3} cal = 3\times 10^{6} cal$$$$T_{1} = 627^{\circ}C = 627 + 273 = 900\ K$$$$T_{2} = 27^{\circ}C = 27 + 273 = 300\ K$$Since $$\dfrac {Q_{1}}{Q_{2}} = \dfrac {T_{1}}{T_{2}}\Rightarrow Q_{2} = \dfrac {T_{2}}{T_{1}} \times Q_{1}$$$$Q_2= \dfrac {300}{900} \times 3\times 10^{6} = 10^{6} cal$$work done $$W = Q_{1} - Q_{2}$$$$= 3\times 10^{6} - 10^{6} = 2\times 10^{6} cal$$$$= 2\times 10^{6} \times 4.2\ J = 8.4\times 10^{6} J$$.Physics

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