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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 ?

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  • $\begingroup$ $5.6(x-y)=y\implies5.6x=6.6y$ $\endgroup$
    – DanDan0101
    Apr 10, 2021 at 17:40
  • $\begingroup$ From my answer do you now see where the 6.6 came from? $\endgroup$
    – Bob D
    Apr 10, 2021 at 19:22

1 Answer 1

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Hint: Start with

$$COP=\frac{\dot Q_C}{{\dot Q_H}-\dot Q_C}$$

You calculated the COP and you are given $\dot Q_H$=10 kw

from there you should get the expression you are questioning.

Hope this helps.

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