Question

a carnot engine absorbs 200J of heat from a reservoir at the temperature of the normal boiling point of water and rejects heat to reservoir at the temperature of the triple point of water . find the heat rejected,the work done by the engine and the thermal efficiency.​

Answer 1

Answer:

The heat rejected is 146J.

The work done by the engine is 54J.

The thermal efficiency is 0.27

Explanation:

The efficiency of the carnot engine, which is working between a hot source at a temperature $T_{1}$ and a cold source at a temperature $T_{2}$, is given by:

$Efficiency=1-\frac{T_{2}}{T_{1}}$

The normal boiling point of water is 373 K. ($T_{1}$)

The temperature of triple point of water is 273.15 K ($T_{2}$)

$Efficiency=1-\frac{273.15}{373}=0.27$

$Efficiency=\frac{Work\:done}{Heat\:absorbed}=\frac{Work\:done}{200}=0.27\\Work\:done=0.27\times200=54J\\Work\:done=Heat\:absorbed-Heat\:rejected\\Heat\:rejected=Heat\:absorbed-Work\:done=200-54=146J$