# Second Law of Thermodynamics - heat pumps and air conditioners

• elcaptain
In summary, a heat pump is operating between 21 degrees Celsius and 41 degrees Celsius, drawing 20000 Btu per hr from a building. The COP is 35% of a Carnot air conditioner. The value of the effective COP is 0.3675. The value of the power required for the compressor motor is 129.25 Watts.
elcaptain

## Homework Statement

A heat pump operates as an air conditioner, drawing 20000 Btu per hr from a building. It is operating between 21 degrees Celsius and 41 degrees Celsius.

a) If the COP (coefficient of performance) is 35% that of a Carnot air conditioner, what is the value of the effective COP (coefficient of performance)?

b) What is the value of the power (Watts) required for the compressor motor?

## Homework Equations

QL = heat removed from the low temperature area (inside the building)
QH = heat removed from the high temperature area (outside the building)
W = work done to remove the heat
TL = temperature of system at lower temperature (inside the building)
TH = temperature of system at higher temperature (outside the building)

COP = QL/W

COP = QL/(QH - QL)

COPideal = TL / (TH - TL)

## The Attempt at a Solution

I first converted Btu (British thermal unit) to joules: 1Btu = 1055.056 J. Then I tried finding the ideal COP by using the above equation with the given temperatures and got 1.05, but I'm pretty sure the ideal COP is not calculated this way. I tried multiplying 1.05 by 0.35 but got the wrong answer for part a. Thanks for any help.

anything?

I would approach this problem by first understanding the Second Law of Thermodynamics. This law states that heat will naturally flow from a hot object to a cold object, and that in order to reverse this flow, work must be done. This is the basis for how heat pumps and air conditioners work.

In this problem, we are given the COP (coefficient of performance) of the heat pump, which is defined as the ratio of the heat removed from the low temperature area (QL) to the work done to remove the heat (W). We are also given the temperatures of the system (TL = 21 degrees Celsius and TH = 41 degrees Celsius).

To solve for the value of the effective COP, we can use the equation COP = QL/W and plug in the given COP (0.35) and the temperature difference (TH-TL = 20 degrees Celsius). This gives us:

0.35 = QL/W
W = QL/0.35

To find the value of QL, we can use the equation COPideal = TL/(TH-TL) and plug in the given temperatures. This gives us:

COPideal = 21/(41-21)
COPideal = 21/20
COPideal = 1.05

Now we can substitute this value into the equation for W:

W = QL/0.35
W = (1.05)/0.35
W = 3 Joules

Therefore, the value of the effective COP is 3 and the power required for the compressor motor is 3 Watts. This means that for every 3 Watts of power used, the heat pump is able to remove 1 Joule of heat from the building.

In conclusion, the Second Law of Thermodynamics explains how heat pumps and air conditioners work by using work to reverse the natural flow of heat. By understanding this law and using the equations for COP and COPideal, we can solve for the value of the effective COP and the power required for the compressor motor.

## 1. What is the Second Law of Thermodynamics?

The Second Law of Thermodynamics states that in any energy conversion process, some energy will be lost or dissipated as heat. This means that the total amount of energy in a closed system will always decrease over time.

## 2. How does a heat pump work?

A heat pump works by transferring heat from a colder area to a warmer area, using a small amount of energy. This is possible because the Second Law of Thermodynamics allows for the transfer of heat from a colder to a warmer area as long as some work is done.

## 3. What is the difference between a heat pump and an air conditioner?

While both heat pumps and air conditioners use the same basic principles of heat transfer, the main difference is their direction of heat flow. A heat pump can both heat and cool a space by transferring heat from one area to another, while an air conditioner can only cool a space by removing heat from the air.

## 4. How efficient are heat pumps and air conditioners?

The efficiency of a heat pump or air conditioner is measured by its Coefficient of Performance (COP). This is the ratio of the amount of heat transferred to the amount of energy used. For most modern heat pumps and air conditioners, the COP ranges from 2 to 4, meaning they are about 200-400% efficient.

## 5. Can a heat pump or air conditioner violate the Second Law of Thermodynamics?

No, a heat pump or air conditioner does not violate the Second Law of Thermodynamics. While they may transfer heat from a colder area to a warmer area, they still require energy to do so and therefore do not create energy out of nothing. The Second Law still holds true as some energy is lost in the process.

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