Question

An ideal heat engine absorbs 10^5 k.cal heat from source at 150°c and rejects the heat to a sink at 20°c.then the work done by the engine is

Answer 1

An ideal heat engine absorbs 10^5 Kcal heat from source at 150°C and rejects the heat to a sink at 20°C.

first of all, we should find efficiency of ideal heat engine, $\eta=1-\frac{T_1}{T_2}$

where $T_1$ temperature of sink in Kelvin and $T_2$ temperature of source in Kelvin.

e.g., $T_1=20+273=293K$

and $T_2=150+273=323K$

now, $\eta=1-\frac{293}{323}$

= 130/323

we also know, efficiency = workdone/heat absorbs from source

or, 130/323 = workdone/10^5 Kcal

or, workdone = 10^5 × 130/323

= 0.402476 × 10^5 KCal

= 40247.6 KCal

hence, workdone by the engine is 40247.6 KCal