A question about refrigerators and energy

[j88k]

Homework Statement

The coefficient of performance of a refrigerator is 5.40. The compressor uses 30.0 J of energy per cycle.

a) How much heat energy is exhausted per cycle?
b) If the hot-reservoir temperature is 27.0C, what is the lowest possible temperature in C of the cold reservoir?

Homework Equations

K=TC/(TH-TC)

QH = QC + Win

The Attempt at a Solution

solved part a) and found the energy to be 192 J

i keep on getting part b) wrong though and I ran out of chances to try.

please help!

 

[Redbelly98]

What is the COP (coefficient of performance) in terms of Qh, Qc, and/or W?

 

[nlightner]

Based on the given information, the coefficient of performance (COP) of the refrigerator can be calculated using the equation COP = QH / Win, where QH is the heat energy exhausted per cycle and Win is the energy input to the compressor. Therefore, using the given COP of 5.40 and energy input of 30.0 J per cycle, we can calculate QH to be 162 J (5.40 x 30.0 J = 162 J). This means that 162 J of heat energy is exhausted per cycle by the refrigerator.

Now, to determine the lowest possible temperature of the cold reservoir, we can use the equation K = TC / (TH - TC), where K is the COP, TC is the temperature of the cold reservoir, and TH is the temperature of the hot reservoir. We can rearrange this equation to solve for TC, giving us the equation TH / (K + 1) = TC. Plugging in the given values of TH = 27.0C and K = 5.40, we get TC = 27.0C / (5.40 + 1) = 27.0C / 6.40 = 4.22C. Therefore, the lowest possible temperature of the cold reservoir is 4.22C.

I hope this helps with solving part b) of the question. Remember to always carefully check your calculations and units to ensure accuracy. Good luck with your homework!