Question

carnots heat engine takes 300J of heat from a source of 627°c and gives some part of it to sink at 27°c. work done by engine in one cycle is​

Answer 1

Explanation:

GIVEN>

$T_{1} = 627 + 273 = 900k \\ T_{2} = 27 + 273 = 300k \\ Q_{1} = 300 = 3 \times {10}^{2}$

According to Carnot theorem

$\frac{ Q_{1} }{T_{1}} = \frac{ Q_{2} }{T_{2}} \\ Q_{2} = \frac{ T_{2} }{T_{1}} \: \times Q_{1} \:$

after substituting value

$Q_{2} = \frac{300}{900} \times 3 \times {10}^{2} \\ = 1 \times {10}^{2}$

Now

$w = Q_{1} - Q_{2} \\ = 3 \times {10}^{2} - 1 \times {10 }^{2} \\ = 2 \times {10}^{2}$

hope it will help you thanks for asking

it can also understand by efficiency method