Question

A heat engine operating between the

Answer 1

Answer:

(Q1 - Q2)/Q1

Explanation:

The efficiency of any device can be given by the following formula:

$Efficiency = \eta$ = Output/Input

Now, for engine the output is the mechanical work that it does and the input is the heat that it absorbs. Let:

W = Work Done by the engine

Q₁ = Heat Absorbed by the System

Q₂ = Heat Rejected by the System (Because According to 2nd Law of Thermodynamics, engine can not convert the entire heat into work without rejection of some heat)

Therefore, the equation becomes:

$\eta = \frac{W}{Q_{1}}$

but, the work can be written as the difference between the absorbed heat and the rejected heat. Hence,

$\eta = \frac{Q_{1} - Q_{2}}{Q_{1}}$

Therefore, the correct option is:

(Q1 - Q2)/Q1