Question

A Carnot engine absorbs 1000 J of heat from a reservoir at 127C and rejects 600 J of heat during each cycle.
Calculate
(a) The efficiency of the engine
(b) The temperature of the sink
(c) The amount of the useful work done during each cycle.

Answer 1

Given:

Heat absorbed, $Q_1=1000J$

Heat rejected, $Q_2=600J$

Temperature of reservoir, $T_1=127°C=127+273=400K$

To find:

Efficiency of the engine, $\eta =?$

Temperature of sink, $T_2=?$

Amount of useful work done per cycle, $W=?$

We know,

Efficiency of Carnot engine can be found be using the following formula,

$\eta =\frac{Q_1-Q_2}{Q_1}=\frac{1000-600}{1000}=0.4\text{or}40\%$

Hence, the efficiency of the engine is 40$\%$

We know, Efficinecy of carnot engine is also equal to

$\eta =\frac{T_1-T_2}{T_1}\therefore 0.4=\frac{400-T_2}{400}⟹400-T_2=160⟹T_2=400-160=240K\text{or}240-273=-33°C$

Hence, the temperature of the sink is -33°C.

We know,

$W=Q_1-Q_2=1000-600=400J$

Hence the amount of useful work done during each cycle is 400J.