An Air Conditioner: Refrigerator or Heat Pump?

[Kalie]

Air conditioners operate on the same principle as refrigerators. Consider an air conditioner that has 7.00 kg of refrigerant flowing through its circuit each cycle. The refrigerant enters the evaporator coils in phase equilibrium, with 54.0% of its mass as liquid and the rest as vapor. It flows through the evaporator at a constant pressure and when it reaches the compressor 95% of its mass is vapor. In each cycle, how much heat Q_c is absorbed by the refrigerant while it is in the evaporator? The heat of vaporization of the refrigerant is 1.50×105 kg/J

All right I don't know what to do
I know I have to find the percentage of refrigerant transformed to vapor which is 49%
Now I have to calculate the heat needed to transform this amount of refrigerant to vapor. But I don't know how, I just need the absic equation please help.

 

[OlderDan]

Find the mass of the liquid that gets converted to to vapor using the percentages. Check the units on the heat of vaporization. Did you type that correctly?

 

[kyle8921]

I can provide a response to this content. First, let's consider the operating principle of an air conditioner. It works by using a refrigerant to absorb heat from the indoor air and transfer it outside, thus cooling the air inside the room. The refrigerant, which is a chemical compound, goes through a continuous cycle of evaporation and condensation to achieve this cooling effect.

Now, let's look at the specific scenario mentioned in the content. We have an air conditioner with 7.00 kg of refrigerant flowing through its circuit each cycle. The refrigerant enters the evaporator coils with 54.0% of its mass as liquid and the rest as vapor. This means that 49% of the refrigerant is already in its vapor state when it enters the evaporator coils.

Next, the refrigerant flows through the evaporator at a constant pressure and when it reaches the compressor, 95% of its mass is vapor. This means that only 5% of the refrigerant is still in its liquid state at this point.

To calculate the heat absorbed by the refrigerant while it is in the evaporator, we need to use the formula Q= m * h. Where Q is the heat absorbed, m is the mass of the refrigerant, and h is the heat of vaporization.

In this case, Q = (7.00 kg * 0.49) * (1.50 x 105 kg/J) = 5.145 x 105 J.

Therefore, in each cycle, the refrigerant absorbs 5.145 x 105 J of heat while it is in the evaporator. This heat is then transferred outside through the condenser and the cycle repeats to cool the indoor air.

I hope this explanation helps you understand the basic equation and how it applies to this scenario. Please let me know if you have any further questions.