Question

a heat engine operates between 20 and 80 degree c calculate max possible efficiency

Answer 1

Answer:

17%

Explanation:

maximum possible efficiency will be in carnot cycle

efficiency =

$1-\frac{T_{L}}{T_{H}}$

take temprature in kelvin

$T_{L} = 20 degree = 273 + 20=293K\\T_{H}=80 degree=273+80=353K\\1 - \frac{T_{L}}{T_{H}} = 1-\frac{293}{353} = 0.17$

Answer 2

Answer:

17 percent

Explanation:

maximum possible efficiency will be in carnot cycle

efficiency =

1-\frac{T_{L}}{T_{H}}

take temprature in kelvin

T_{L} = 20 degree = 273 + 20=293K\\T_{H}=80 degree=273+80=353K\\1 - \frac{T_{L}}{T_{H}} = 1-\frac{293}{353} = 0.17