# At what rate does the air conditioner remove energy from the laboratory?

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 ?

• $5.6(x-y)=y\implies5.6x=6.6y$ Apr 10, 2021 at 17:40
• From my answer do you now see where the 6.6 came from? Apr 10, 2021 at 19:22

$$COP=\frac{\dot Q_C}{{\dot Q_H}-\dot Q_C}$$
You calculated the COP and you are given $$\dot Q_H$$=10 kw