Answer to Question #295606 in Molecular Physics | Thermodynamics for Marww

Question

The energy absorbed by an engine is two times greater than the work it performs.

What is its thermal efficiency?

What fraction of the energy absorbed is expelled to the cold reservoir?

Answer 1

Given :$|Q_h|$ (energy absorbed by the engine= $2W_{eng}$

a) the thermal efficiency of the engine is given by $e=\frac{W_{eng}}{|Q_h|}$

Substitute for $|Q_h|$ from given data

$e=\frac{W_{eng}}{2W_{eng}}=\frac{1}{2}$

(b) The energy $|Q_c|$ expelled to the cold reservoir is found using

$∣Q_c∣=∣Q_h∣−W_{eng} ​$

Substitute for $|Qh​∣$ from given data:

$∣Q _c ​ ∣=2W_{eng}−W_{eng}=W_{ eng} ​$

Then, the ratio of energy$∣Q_h∣$ absorbed by the engine from the hot reservoir to the energy$∣Q_c|$

expelled by the engine to the cold reservoir is:

$e=\frac{2W_{eng}}{1W_{eng}}=2$