Thermodynamics: Coefficient of Performance Equation for Temperature Calculation

[name_ask17]

Just a quick question. Is there a formula for the coefficient of performance equation for just temperatures. All I have is the outdoor and indoor temperauture. How would I approach that?

 

[collinsmark]

Just a quick question. Is there a formula for the coefficient of performance equation for just temperatures. All I have is the outdoor and indoor temperauture. How would I approach that?

Yes, there is such a formula. It generally assumes that the heat pump (or refrigerator) is an ideal Carnot engine (maximum efficiency -- no losses due to friction, etc.)

You should be able to find this set of equations in your textbook or coursework. There is one COP equation for heating (such as what is used by a heat pump) and another, similar equation for cooling (such as what is used by a refrigerator or air conditioner).

Or you can derive it yourself. For an ideal system, Q_hot/T_hot = Q_cold/T_cold.

 

[name_ask17]

what is Q? how would i find that.
because in my problem, i literally only have Tc and Th. how do i do that?

 

[collinsmark]

what is Q? how would i find that.
because in my problem, i literally only have Tc and Th. how do i do that?

Do you already have formulas for COP as a function of Q_hot and Q_cold?

Make your substitutions. For example, Q_cold = T_coldQ_hot/T_hot, for an ideal Carnot engine. Make a similar substitution for Q_hot.

 

[name_ask17]

ok. this is my problem.
In a heat pump, heat from the outdoors at -8.00 deg c is transferred to a room at 22 deg c, energy being supplied by an electrical motor. How mnay joules of heat will be delivered to the room for each joule of electrical energy consumed, ideally?

when it asks for joules of heat, what specifically is this asking me to solve for? and if i use the equation you wrote above, do i have to convert it to kelvin or can i leave it as celsius?

 

[collinsmark]

when it asks for joules of heat, what specifically is this asking me to solve for?

The coefficient of performance or COP of a heat pump is the ratio of the change in heat at the "output" (the heat reservoir of interest) to the supplied work.

So, the question is asking for just that. The question is asking for the COP of the system, assuming it is ideal.

And the question is asking for a ratio. So your answer is not going to be in units of Joules. Rather it's going to be a unit-less number. COP is a unit-less number.

and if i use the equation you wrote above, do i have to convert it to kelvin or can i leave it as celsius?

You absolutely have to convert to Kevlin!

You'll also at least have to find a formula for the COP in your textbook. I've given you enough information where you can convert the formula to be a function of T_hot and T_cold instead of Q_hot and Q_cold. But you'll at least need to find a formula for your textbook or coursework that gives you the COP in terms of Q_hot and Q_cold.