A biology laboratory is maintained at a constant temperature of $7.00°C$ by an air conditioner, which is vented to the air outside. On a typical hot summer day the outside temperature is $27.0°C$ and the air conditioning unit emits energy to the outside at a rate of $10.0$ kW. Model the unit as having a coefficient of performance equal to $40.0$% of the coefficient of performance of an ideal Carnot device.
$(a)$ At what rate does the air conditioner remove energy from the laboratory?
Like a refrigerator, an air conditioner has as its purpose the removal of energy by heat from the cold reservoir.
Its ideal COP is $COP_{Carnot}=\frac{T_{c}}{T_{h}-T_{c}}$
but
$T_{c}=7°C=280°K$ and $T_{h}=27°C=300°K$
Then
$COP_{Carnot}=\frac{280°K}{20°K}=14.0$
Its actual COP is
$(0.400)(14.0)=5.60=\frac{|Q_{c}|}{|Q_{h}|-|Q_{c}|}=\frac{|\frac{Q_{c}}{\Delta t}|}{|\frac{Q_{h}}{\Delta t}|-|\frac{Q_{c}}{\Delta t}|}$
$5.60(|\frac{Q_{h}}{\Delta t}|-|\frac{Q_{c}}{\Delta t}|)=|\frac{Q_{c}}{\Delta t}|$
The next step is where it confuses me as it does the following
$5.60(10.0kW)=6.60|\frac{Q_{c}}{\Delta t}|$
And I can't understand where the $6.60$ comes from. By definition we have the following
$W=|Q_{h}|-|Q_{c}|$ and $P=\frac{W}{\Delta t}$
But it causes me a lot of confusion because I can see where the power comes from $P = 10.0kW$ but the term $6.60$ does not. Could you help me ?
