Question

A heat engine operates between 20°c and 80°c calculate maximum possible efficiency of this engine

Answer 1

Answer:

Explanation:

Eff.= 1-T2÷T1 × 100

T1=80 & T2=20

=> 1-20÷80 × 100

=> 0.75 × 100 = 75% efficiency

Answer 2

A heat engine operates between $20^{0} and 80^{0}$ so, the maximum possible efficiency of this engine is 16.9%

Explanation:

Let,

$T_{1} = 20^{0}C\\ \\T_{2} = 80^{0}C$

To find,

Maximum efficiency of engine.

Temperature from celsius to kelvin,

$T_{1} = 20 + 273 = 293K\\ \\T_{2} = 80 + 273 = 353K$

The ratio of useful work done to the heat provided is an efficiency of heat engine.

Formula used

η = $\frac{T_{2}-T_{1} }{T_{2} }$

η = $\frac{353-293}{353}$

η =$\frac{60}{353}$

η = 0.1699

Therefore, the maximum possible efficiency of this engine is 16.9%.