Easy thermodynamics problem

In summary, the problem discussed is about determining the power required to operate a heat pump in order to maintain a house at 24°C when the outdoor temperature is -5°C and the house is losing heat at a rate of 80,000 kJ/h. The conversation also mentions using equations such as COP = Qh/(Qh-Ql) = Qh/W, and discussing the use of a Carnot cycle in solving the problem. However, it is unclear what type of thermodynamic cycle is being used in the problem and it is suggested to assume a Carnot cycle for solving it. The final answer for the power required is 2.17 kW.
  • #1
Physics_wiz
228
0
This problem is supposedly easy but I just can't figure out how to do it.

A heat pump is used to heat a house and maintain it at 24°C. On a winter day when the outdoor temperature is -5°C, the house is estimated to lose heat at a rate of 80,000 kJ/h. Determine the power required to operate this heat pump

Only two equations I know are: COP = Qh/(Qh-Ql) = Qh/W
where W is the net work done (what I'm looking for).

In the book, they have an example just like it except the COP is given so all I got to do is plug numbers in the equation and get the answer but I don't know how to do it when the temperatures are given instead of the COP.
 
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  • #2
Physics_wiz said:
This problem is supposedly easy but I just can't figure out how to do it.

A heat pump is used to heat a house and maintain it at 24°C. On a winter day when the outdoor temperature is -5°C, the house is estimated to lose heat at a rate of 80,000 kJ/h. Determine the power required to operate this heat pump

Only two equations I know are: COP = Qh/(Qh-Ql) = Qh/W
where W is the net work done (what I'm looking for).

In the book, they have an example just like it except the COP is given so all I got to do is plug numbers in the equation and get the answer but I don't know how to do it when the temperatures are given instead of the COP.
Does the question also say that the heat pump operates on an ideal Carnot cycle? If so:

[tex]\Delta S = Q_H/T_H - Q_C/T_C = 0[/tex] so:

[tex]Q_H/T_H = Q_C/T_C \rightarrow Q_H/Q_C = T_H/T_C[/tex]

So [tex]COP = \frac{T_H}{T_H - T_C}[/tex]

AM
 
  • #3
No, if it were a carnot I'd know how to do it but it doesn't say anything about that in the problem so I don't think I can just assume carnot cycle.
 
  • #4
I think I'm wasting too much time on this problem and I have a test tomorrow. Anyone know if there's a way to solve this problem if it were not a carnot cycle?

Thanks.
 
  • #5
Physics_wiz said:
I think I'm wasting too much time on this problem and I have a test tomorrow. Anyone know if there's a way to solve this problem if it were not a carnot cycle?
Not possible. It would depend on the details of the thermodynamic cycle used. Without knowing that, you cannot answer the question. I would suggest that you assume a Carnot cycle.

AM
 
  • #6
Alright. So, assuming it's a carnot cycle, this is how it should be solved:

80,000 KJ/h = 2.22 KW

COP = Th/(Th-Tc) = 297 k/(297 k-268 k) = 10.24

COP = Qh/W ---> W = Qh/COP

W = 2.22 KW/10.24 = 2.17 KW which should be equal to the power needed to operate the pump.

Did I do anything wrong?
 
  • #7
Physics_wiz said:
Alright. So, assuming it's a carnot cycle, this is how it should be solved:

80,000 KJ/h = 2.22 KW

COP = Th/(Th-Tc) = 297 k/(297 k-268 k) = 10.24

COP = Qh/W ---> W = Qh/COP

W = 2.22 KW/10.24 = 2.17 KW which should be equal to the power needed to operate the pump.

Did I do anything wrong?
Just that Qh = 22.2 kW but your answer is right.

AM
 
  • #8
Yes, that is right except as Andrew explained.
 
  • #9
Aight thanks. You guys said that Qh isn't equal to 22.2 KW because that would be Qh/time. The units for Qh should be in KJ or J's I guess right?
 
  • #10
No, Qh is indeed 22.2kW but you said 2.22kW
 
  • #11
oh...I should probably go to bed now because my internal energy is going up, info in my head is starting to vaporize, my brain is becoming superheated, volume not changing since my skull is rigid, pressure increasing, enthalpy going up, aaaand I need to get some rest to cool down and condense the info again so it's usable by the time of the test. :zzz:
 

1. What is thermodynamics?

Thermodynamics is the branch of physics that deals with the relationships between heat, work, temperature, and energy.

2. What is an easy thermodynamics problem?

An easy thermodynamics problem is typically a problem that involves applying basic principles and equations to solve for a specific variable, such as temperature or heat transfer.

3. How do I solve an easy thermodynamics problem?

To solve an easy thermodynamics problem, you will need to identify the known variables, the unknown variable, and the relevant equations. Then, you can plug in the known values and solve for the unknown variable using algebraic manipulation.

4. What are some common equations used in easy thermodynamics problems?

Some common equations used in easy thermodynamics problems include the first and second laws of thermodynamics, the ideal gas law, and the specific heat capacity equation.

5. What are some tips for solving an easy thermodynamics problem?

Some tips for solving an easy thermodynamics problem include drawing a diagram to visualize the problem, checking the units of all variables, and double-checking your calculations for accuracy.

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