Calculating COP for Refrigerators and Heat Pumps

1. Refrigerators and Heat Pumps: Heat Engines in Reverse

A heat engine absorbs heat $|Q_H|$ from a high-temperature reservoir ($T_H$), converts part of it into net work $W$, and rejects the rest, $|Q_C|$, to a low-temperature reservoir ($T_C$).

A refrigerator or heat pump does the opposite. It uses an input of work $W$ to extract heat $|Q_C|$ from a cold reservoir and reject a larger amount of heat $|Q_H|$ to a hot reservoir. This is why the back of your refrigerator feels warm; it's dumping heat from inside the fridge (and the heat generated by the work done) into your kitchen.

To get a clear conceptual picture of this reversal, let's watch a short introductory video.

The core principle, as always, is energy conservation (the First Law of Thermodynamics). For one complete cycle, the change in internal energy of the working substance is zero ($\Delta U=0$). This means the total energy going into the device must equal the total energy coming out.

Looking at the device (the circle in the diagram below):

• Energy In: Work $W$ and heat from the cold reservoir $Q_C$.
• Energy Out: Heat rejected to the hot reservoir $Q_H$.

Therefore, for a full cycle, we have:
$|Q_H|=|Q_C|+|W|$

Note on Sign Conventions: To avoid confusion, especially since you've requested clarity on this, we will use the magnitudes of heat ($|Q|$) and work ($|W|$) for these cycle calculations. In the context of COP, $W$ is the work input required to run the device, $Q_C$ is the heat extracted from the cold space, and $Q_H$ is the heat rejected to the hot space. All are treated as positive quantities in the COP formulas.

The following diagram provides an excellent visual summary comparing a heat engine with a refrigerator and a heat pump.

The diagram reveals a subtle but critical distinction:

• A refrigerator's purpose is to cool the cold space. Its "desired output" or "benefit" is the heat it removes, $|Q_C|$.
• A heat pump's purpose is to heat the hot space. Its "desired output" is the heat it delivers, $|Q_H|$. An air conditioner in winter is acting as a heat pump.

2. Coefficient of Performance (COP)

For these devices, "efficiency" is not the right term because work is an input (a "cost"), not an output. Instead, we use the Coefficient of Performance (COP), often denoted by $K$, $\beta$, or simply COP. It's defined as:
$\text{COP}=\frac{\text{Desired Output}}{\text{Required Input}}=\frac{\text{Benefit}}{\text{Cost}}$

COP of a Refrigerator ($K_R$)

The benefit is the heat extracted from the cold reservoir, and the cost is the work done.
$K_R=\frac{|Q_C|}{|W|}$
Using our energy balance equation $|W|=|Q_H|-|Q_C|$, we can also write this as:
$K_R=\frac{|Q_C|}{|Q_H|-|Q_C|}$

COP of a Heat Pump ($K_P$)

The benefit is the heat delivered to the hot reservoir, and the cost is the same work input.
$K_P=\frac{|Q_H|}{|W|}$
Similarly, this can be written as:
$K_P=\frac{|Q_H|}{|Q_H|-|Q_C|}$
Notice that COP values are typically greater than 1. A COP of 4 means that for every 1 Joule of work you put in, the device moves 4 Joules of heat.

3. A Key Relationship: $K_P$ vs. $K_R$

For the same device operating between the same two temperatures, there's a simple and powerful relationship between its performance as a heat pump and as a refrigerator.

Starting with the definition of $K_P$:
$K_P=\frac{|Q_H|}{|W|}$
Substitute $|Q_H|=|Q_C|+|W|$:
$K_P=\frac{|Q_C|+|W|}{|W|}=\frac{|Q_C|}{|W|}+\frac{|W|}{|W|}$
Recognizing that $\frac{|Q_C|}{|W|}=K_R$, we get:
$K_P=K_R+1$
This elegant result means that a heat pump's COP is always exactly one greater than the COP of a refrigerator operating under the same conditions. This makes sense: the heat delivered to the hot side is the sum of the heat taken from the cold side plus the work done.

4. Ideal Performance: The Carnot Refrigerator/Heat Pump

Just as the Carnot cycle sets the maximum possible efficiency for a heat engine, a reversed Carnot cycle sets the maximum possible COP for a refrigerator or heat pump operating between two temperatures, $T_H$ and $T_C$.

For a reversible (Carnot) cycle, we have the relation:
$\frac{|Q_H|}{|Q_C|}=\frac{T_H}{T_C}$
(Remember, temperatures must be in Kelvin).

We can substitute this into our COP formulas to find the maximum theoretical performance.

Maximum COP of a Refrigerator

$(K_R{)}_{\text{max}}=\frac{|Q_C|}{|Q_H|-|Q_C|}=\frac{1}{\frac{|Q_H|}{|Q_C|}-1}=\frac{1}{\frac{T_H}{T_C}-1}$
$(K_R{)}_{\text{max}}=\frac{T_C}{T_H-T_C}$

Maximum COP of a Heat Pump

$(K_P{)}_{\text{max}}=\frac{|Q_H|}{|Q_H|-|Q_C|}=\frac{\frac{|Q_H|}{|Q_C|}}{\frac{|Q_H|}{|Q_C|}-1}=\frac{\frac{T_H}{T_C}}{\frac{T_H}{T_C}-1}$
$(K_P{)}_{\text{max}}=\frac{T_H}{T_H-T_C}$

These formulas are crucial. They tell you the best-case-scenario performance based only on the operating temperatures. Any real device will have a COP lower than this Carnot value due to irreversibilities like friction.

5. Application and Problem Solving

Let's see how these formulas are applied in a typical problem.

Now, it's your turn to apply these concepts. This problem structure is very common in JEE questions.

Conclusion

In this lesson, you've moved from heat engines to heat pumps and refrigerators, learning how to quantify their performance.

Key Takeaways:

• Refrigerators and heat pumps are heat engines operating in reverse; they use work $W$ to move heat from a cold reservoir ($Q_C$) to a hot one ($Q_H$).
• The Coefficient of Performance (COP) measures their effectiveness:
  • Refrigerator: $K_R=|Q_C|/|W|$ (focus on cooling)
  • Heat Pump: $K_P=|Q_H|/|W|$ (focus on heating)
• For any device, the two COPs are related by $K_P=K_R+1$.
• The maximum possible COP is achieved by a reversible Carnot cycle and depends only on the reservoir temperatures (in Kelvin):
  • $(K_R{)}_{\text{max}}=T_C/(T_H-T_C)$
  • $(K_P{)}_{\text{max}}=T_H/(T_H-T_C)$

Preview of the Next Lesson:
We've now fully explored the First Law's application to cyclic devices and have been introduced to the limitations imposed by the Second Law (e.g., Carnot efficiency/COP). To delve deeper into the Second Law, we need a new concept: entropy. In the next lesson, you will learn how to define entropy and its relation to the second law of thermodynamics, giving you a powerful tool to determine the direction and possibility of thermodynamic processes.