Thermodynamics, Refrigerator

[theown1]

Homework Statement

An air conditioner operates on 800W of power and has a performance coefficient of 2.80 with a room temperature of 21^oC, and an outside temperature of 35^oC.
a) Calculate the rate of heat removal for this unit.
b) calculate the rate at which heat is discharged to the outside air.
c) calculate the total entropy change in the room if the air conditioner runs for 1 hour. Calculate the total entropy change in the outside air for the same time period.
d) What is the net change in entropy for the system(room+outside air)?

Homework Equations

$$\Delta$$Q=cm$$\Delta$$T
I think that that is the equation of heat lost
Coefficient of performance = Q_{cold, input}/W_input
C.o.P=Q_cold,input/(Q_hot,output-Q_cold,input)

The Attempt at a Solution

a) I'm not sure how to find the rate of heat which is removed

b) I'm not sure either

c) I know that Entropy=Q/T, and
$$\Delta$$S=mcln(T_h/T_c)
c is the specific heat, m is mass

but I'm not sure how to find the final temperature from what's given in the problem, or the mass...


d) once I know the answer to part C, its fairly self explanatory

Can someone guide in the right direction for solving this

 

[soothsayer]

for part a, the useful equations are CoP= Q_ci/(Q_ho-Q_ci and CoP = Q_ci/W. However, since we only have a value for power, W/t, to use the latter equation, you have to turn everything into a rate. This isn't hard to do, just multiply the numerator and denominator by 1/t, you'll get CoP = (Q_ci/t)/P. Now plug in the known values for P and CoP, can you take it from there? (hint, you'll have to use the same trick for the equation involving Q_ho

 

[theown1]

Ok thanks that's makes a little more sense now, thanks

 

[soothsayer]

No problem, that should give you the answer for a and b. For part c, remember,

This should help you find the temperature change.

 

[rwisz]

problem?

a) The rate of heat removal for the air conditioner can be calculated using the equation Q = mCΔT, where Q is the rate of heat removal, m is the mass of the air being cooled, C is the specific heat of air, and ΔT is the temperature difference between the room and the air being discharged. In this case, we can assume a standard room size and use the specific heat of air at constant pressure, which is approximately 1 kJ/kg·K. The temperature difference would be 35oC - 21oC = 14oC. Therefore, the rate of heat removal for the air conditioner would be Q = m(1 kJ/kg·K)(14oC) = 14m kJ/s.

b) The rate at which heat is discharged to the outside air can be calculated using the equation Q = mCΔT, where Q is the rate of heat discharge, m is the mass of the air being discharged, C is the specific heat of air, and ΔT is the temperature difference between the air being discharged and the outside air. In this case, we can again assume a standard room size and use the specific heat of air at constant pressure. The temperature difference would be 35oC - 21oC = 14oC. Therefore, the rate of heat discharge to the outside air would be Q = m(1 kJ/kg·K)(14oC) = 14m kJ/s.

c) The total entropy change in the room can be calculated using the equation ΔS = Q/T, where ΔS is the total entropy change, Q is the heat removed from the room, and T is the room temperature. From part a, we know that Q = 14m kJ/s. The room temperature is 21oC, which is equivalent to 294 K. Therefore, the total entropy change in the room would be ΔS = (14m kJ/s)/294 K = 0.048m kJ/K.

To calculate the total entropy change in the outside air, we can use the same equation but with the heat discharged to the outside air and the outside temperature. From part b, we know that Q = 14m kJ/s and the outside temperature is 35oC, which is equivalent to 308 K. Therefore, the total entropy change in the outside air would be ΔS