Question

During each cycle, a heat engine operating between two heat reservoirs absorbs 156 J from the reservoir at 100°C and releases 136 J to the reservoir at 20°C.

(a) What is the efficiency of this engine? %

(b) What is the ratio of its efficiency to that of a Carnot engine working between the same reservoirs? This ratio is called the second law efficiency. εengine / εCarnot =

Answer 1

The efficiency of the engine is

$\eta= 1-\frac{Q_2}{Q_1} \Rightarrow 1 - \frac{136}{156} = 0.128 = 12.8 \%$

Efficiency of carnot engine is

$\eta= 1-\frac{T_2}{T_1} \Rightarrow 1 - \frac{20}{100} = 0.8 = 80 \%$

The ratio of the efficiencies is

$\frac{n_e}{n_c} = \frac{0.128}{0.8} = 16 \%$