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Standard XII

Physics

2nd Law of Thermodynamics

A system unde...

Question

A system undergoes a cyclic process in which it absorbs $Q_{1}$ heat and gives out $Q_{2}$ heat. The efficiency of the process is $η$ and the work done is $W$ .

W = Q_{1} - Q_{2}

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η = \frac{W}{Q_{1}}

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η = \frac{Q_{2}}{Q_{1}}

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η = 1 - \frac{Q_{2}}{Q_{1}}

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Solution

The correct options are
A $η = \frac{W}{Q_{1}}$
B $W = Q_{1} - Q_{2}$
D $η = 1 - \frac{Q_{2}}{Q_{1}}$
The efficiency of a heat engine is defined as the ratio of work output to the heat input.
Thus, $η = \frac{W}{Q_{i n}}$
Now, since the heat energy only gets converted into work and no other form of energy, thus following the law of conservation of energy,
$W = Q_{o u t} - Q_{i n}$
Substituting this value in the above expression, we get another expression for the efficiency of a heat engine

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A system undergoes a cyclic process in which it absorbs Q1 heat and gives out Q2 heat. The efficiency of the process is η and the work done is W.

A system undergoes a cyclic process in which it absorbs $Q_{1}$ heat and gives out $Q_{2}$ heat. The efficiency of the process is $η$ and the work done is $W$ .