Carnot refrigerator and work done

[sportsrules]

The question reads...
A heat pump is essentially a refrigerator engine that uses the inside of a building as a hot reservoir in the winter and the outside of the building as the hot reservoir in the summer. A) For a carnot refrigerator engine operating between 25 C and 40 C in the summer, what is the coefficient of performance? B) For each joule of heat transfer form the cooler reservoir per cycle, how many joules of work are done by the carnot refrigerator engine?

Ok, I figured out part A, which was simply plugging the numbers into the equation

K=1/[(Thot/Tcold)-1]...and I got that K=19.9

I can't really figure out part B. I know that there is another equation

K=Qc/W
and
K=1/[(Qh/Qc)-1]

but I cannot figure out how to rearrange get an answer for the work b/c I don't know what Qc is or how to go about finding it. Any help would be great! Thanks!

 

[Andrew Mason]

The question reads...
A heat pump is essentially a refrigerator engine that uses the inside of a building as a hot reservoir in the winter and the outside of the building as the hot reservoir in the summer. A) For a carnot refrigerator engine operating between 25 C and 40 C in the summer, what is the coefficient of performance? B) For each joule of heat transfer form the cooler reservoir per cycle, how many joules of work are done by the carnot refrigerator engine?

For a heat pump, the coeff. of perf. is the ratio of heat output (heat transferred from the cold reservoir to the hot reservoir) to the work input.

$$COP = \frac{Q_H}{W}$$

So:

$$W = Q_H/COP$$

$$COP = Q_H/W = Q_H/Q_H-Q_C = \frac{1}{1-\frac{Q_C}{Q_H}}$$

For the Carnot process (reversible so $\Delta S = 0$), the heat is exchanged at constant temperature, so:

$$\Delta S = Q_h/T_h - Q_c/T_c = 0$$

$$Q_h/T_h = Q_c/T_c$$

$$Q_h/Q_c = T_h/T_c$$

So: $$COP = \frac{1}{1-\frac{T_C}{T_H}}$$

COP = 1/(1-298/313) = 20.86

So for each Joule of heat delivered, W = 1/20.86 = .048 Joules of work input are needed

AM

 

[thepatientmental]

A) The coefficient of performance (K) for a Carnot refrigerator engine operating between 25 C and 40 C in the summer is calculated using the equation K=1/[(Thot/Tcold)-1]. Plugging in the values, we get K=19.9. This means that for every joule of work input, the refrigerator can transfer 19.9 joules of heat from the cold reservoir to the hot reservoir.

B) To find the work done by the refrigerator engine, we can use the equation K=Qc/W, where Qc is the heat transferred from the cold reservoir and W is the work done. Rearranging this equation, we get W=Qc/K. We already know that K=19.9, so now we just need to find Qc.

To find Qc, we can use the equation K=1/[(Qh/Qc)-1]. Rearranging this equation, we get Qc=Qh/(K+1). We know that Qh is the heat transferred from the hot reservoir, which is equal to the work input. So, Qh=W. Plugging this into the equation for Qc, we get Qc=W/(K+1).

Now, we can plug in the values for W and K to find Qc. W is equal to 1 joule (since we are looking at the work done per joule of heat transfer from the cold reservoir) and K is 19.9. Therefore, Qc=1/(19.9+1)=0.0499 joules.

So, for every joule of heat transfer from the cold reservoir, the Carnot refrigerator engine does 0.0499 joules of work.