Question

A cyclic heat engine does 50kj of work per cycle.If efficiency of engine is 75%,then calculate the heat rejected per cycle.

Answer 1

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Explanation:

Carnot efficiency = Work done/Heat supplied(Q1)

0.75 = 50/ Q1

or, Q1 = 200/3

and, Work done = Q1– Q2

or, Q2 = 200/3-50 = 50/3 = 16.6kJ.

Answer 2

Given that,

Work done by a cyclic heat engine, W = 50 kJ

The  efficiency of engine is 75 %

To find,

The heat rejected per cycle.

Solution,

The efficiency of Carnot heat engine is given by work done divided by heat supplied. So,

$\eta=\dfrac{W}{Q_1}\\\\0.75=\dfrac{50}{Q_1}\\\\Q_1=\dfrac{200}{3}$

We know that, work done is given by :

$W=Q_1-Q_2$

$Q_2$ is heat rejected per cycle.

$Q_2=Q_1-W\\\\Q_2=\dfrac{200}{3}-50\\\\Q_2=16.66\ kJ$

So, the heat rejected per cycle. is 16.67 kJ.

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