Question

the co efficient of performance of air conditionar which absorbs heat from cooling coil at 10 C and expels heat to outside at 40degree C will be
a)10.4
b)9.43
c)-10.4
d)0.33
solve..

Answer 1

@AllTehMaffs ..need ur help .. one more question is out of course .Can u help me ? :)

Answer 2

answer key shows D is correct ..No idea how ??

Answer 3

It's a backwards heat engine. Normally, a heat engine would be powered by heat from the Hot reservoir doing work and then being expelled into the cold reservoir.

First law of Thermodynamics says that energy is conserved, so the engine (backwards engine in our case) is a closed system of energy. All of the QH supplied has to be accounted for by work done, W, and heat expelled QC. Work is extracted from heat flowing from the hot reservoir into the cold reservoir.


\[ Q_h =Qc + W\]
\[ W = Qh - Qc\]

We can look at ours like this, with the arrows reversed.


Heat is being taken away from the cold reservoir by having work done on it, and excess heat being ejected into the heat reservoir. So instead of both reservoirs going to equilibrium, one stays cold and one stays hot through work done on the system.

Efficiency is always a cost benefit ration, and in our case, called the cost of performance, the benefit is the QC, the heat being taken away, versus the work done on the system.

\[ COP = \frac{Qc}{W} = \frac{Qc}{Qh - Qc} = \frac{1}{\frac{Qh}{Qc}-1}\]

Second law of thermo. says that the entropy dumped into the hot reservoir must be at least as much as the entropy absorbed from the cold reservoir. Using that the change in entropy is related to the heat input by

** During quasistatic processes**
\[ \Delta S = \frac{Q}{T} \]

The second law says
\[\frac{Q_h}{T_h} ≥ \frac{Q_c}{T_c} \ \ \ ,\ \ \ \frac{Q_h}{Q_c} ≥ \frac{T_h}{T_c} \]
Substituting....
\[ COP ≤ \frac{1}{\frac {T_h}{T_c}-1} = \frac{T_c}{T_h - T_c} = \frac{10ºC}{30ºC} = \frac{1}{3} \\ COP ≤ \frac{1}{3}\]

The answer is D. Your questions are getting steadily more complicated! :P :)

** Borrowed heavily from Thermal Physics, Daniel V. Schroeder**