The COP Formula, or Coefficient of Performance formula, measures how efficiently a refrigerator or heat pump converts work input into useful heat transfer, expressed as \ ( COP = Q/W \). This concept is covered in NCERT Class 11 Physics, Chapter 12 (Thermodynamics), and is a high-weightage topic for both CBSE board exams and JEE/NEET competitive exams. In this article, you will find the complete COP formula, its derivation, a comprehensive formula sheet, three solved examples at progressive difficulty levels, and expert exam tips for 2025-26.
Key COP Formulas at a Glance
Quick reference for the most important Coefficient of Performance formulas.
- COP of Refrigerator: \( COP_R = \frac{Q_L}{W} \)
- COP of Heat Pump: \( COP_{HP} = \frac{Q_H}{W} \)
- Work Input: \( W = Q_H – Q_L \)
- Relation between COP and Heat Pump: \( COP_{HP} = COP_R + 1 \)
- Carnot COP (Refrigerator): \( COP_{Carnot} = \frac{T_L}{T_H – T_L} \)
- Carnot COP (Heat Pump): \( COP_{HP,Carnot} = \frac{T_H}{T_H – T_L} \)
- Efficiency of Heat Engine: \( \eta = 1 – \frac{T_L}{T_H} \)
What is the COP Formula?
The COP Formula defines the Coefficient of Performance, a dimensionless ratio that quantifies the efficiency of a refrigerator, air conditioner, or heat pump. Unlike a heat engine that converts heat into work, a refrigerator uses work to transfer heat from a cold reservoir to a hot reservoir. The COP formula tells us how much useful heat transfer is achieved per unit of work supplied.
In NCERT Class 11 Physics, Chapter 12 (Thermodynamics), the COP is introduced alongside Carnot’s theorem and the second law of thermodynamics. A higher COP value indicates a more efficient device. For a refrigerator, the COP is the ratio of heat removed from the cold reservoir to the net work done on the system. For a heat pump, the COP is the ratio of heat delivered to the hot reservoir to the net work done.
The COP formula is dimensionless because both the numerator and denominator have units of energy (Joules). Real-world refrigerators have COP values typically between 2 and 6, while Carnot refrigerators set the theoretical upper limit for any given pair of temperatures.
COP Formula — Expression and Variables
The standard COP Formula for a refrigerator is:
\[ COP_R = \frac{Q_L}{W} = \frac{Q_L}{Q_H – Q_L} \]
The COP Formula for a heat pump is:
\[ COP_{HP} = \frac{Q_H}{W} = \frac{Q_H}{Q_H – Q_L} \]
The Carnot COP Formula for a refrigerator (maximum theoretical efficiency) is:
\[ COP_{Carnot} = \frac{T_L}{T_H – T_L} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( COP_R \) | Coefficient of Performance of Refrigerator | Dimensionless |
| \( COP_{HP} \) | Coefficient of Performance of Heat Pump | Dimensionless |
| \( Q_L \) | Heat absorbed from cold reservoir | Joule (J) |
| \( Q_H \) | Heat rejected to hot reservoir | Joule (J) |
| \( W \) | Net work input to the system | Joule (J) |
| \( T_L \) | Temperature of cold reservoir | Kelvin (K) |
| \( T_H \) | Temperature of hot reservoir | Kelvin (K) |
Derivation of the COP Formula
The derivation of the COP formula follows directly from the first law of thermodynamics applied to a refrigeration cycle.
Step 1: A refrigerator absorbs heat \( Q_L \) from the cold reservoir (inside the fridge) and rejects heat \( Q_H \) to the hot reservoir (the room).
Step 2: By energy conservation (first law), the net work input equals the difference in heat flows: \( W = Q_H – Q_L \).
Step 3: The COP is defined as useful output divided by work input. For a refrigerator, the useful output is \( Q_L \). Therefore: \( COP_R = Q_L / W = Q_L / (Q_H – Q_L) \).
Step 4: For a Carnot refrigerator, heat transfer is proportional to absolute temperature: \( Q_L / Q_H = T_L / T_H \). Substituting gives the Carnot COP: \( COP_{Carnot} = T_L / (T_H – T_L) \).
Complete Thermodynamics Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| COP of Refrigerator | \( COP_R = Q_L / W \) | Q_L = heat from cold reservoir, W = work input | Dimensionless | Class 11, Ch 12 |
| COP of Heat Pump | \( COP_{HP} = Q_H / W \) | Q_H = heat to hot reservoir, W = work input | Dimensionless | Class 11, Ch 12 |
| Relation: Heat Pump and Refrigerator | \( COP_{HP} = COP_R + 1 \) | COP_HP = heat pump COP, COP_R = refrigerator COP | Dimensionless | Class 11, Ch 12 |
| Carnot COP (Refrigerator) | \( COP_{Carnot} = T_L / (T_H – T_L) \) | T_L = cold temp (K), T_H = hot temp (K) | Dimensionless | Class 11, Ch 12 |
| Carnot COP (Heat Pump) | \( COP_{HP,Carnot} = T_H / (T_H – T_L) \) | T_H = hot temp (K), T_L = cold temp (K) | Dimensionless | Class 11, Ch 12 |
| Carnot Efficiency (Heat Engine) | \( \eta = 1 – T_L / T_H \) | T_L = sink temp (K), T_H = source temp (K) | Dimensionless (or %) | Class 11, Ch 12 |
| First Law of Thermodynamics | \( \Delta U = Q – W \) | ΔU = internal energy change, Q = heat, W = work | Joule (J) | Class 11, Ch 12 |
| Work Done in Isothermal Process | \( W = nRT \ln(V_2/V_1) \) | n = moles, R = gas constant, T = temperature | Joule (J) | Class 11, Ch 12 |
| Entropy Change | \( \Delta S = Q_{rev} / T \) | Q_rev = reversible heat, T = temperature (K) | J/K | Class 11, Ch 12 |
| Work Input (Refrigerator) | \( W = Q_H – Q_L \) | Q_H = heat rejected, Q_L = heat absorbed | Joule (J) | Class 11, Ch 12 |
COP Formula — Solved Examples
Example 1 (Class 11 Level — Direct Application)
Problem: A refrigerator absorbs 200 J of heat from its cold compartment and the compressor does 80 J of work in one cycle. Calculate the COP of the refrigerator and the heat rejected to the surroundings.
Given: \( Q_L = 200 \) J, \( W = 80 \) J
Step 1: Write the COP formula for a refrigerator: \( COP_R = \frac{Q_L}{W} \)
Step 2: Substitute the values: \( COP_R = \frac{200}{80} = 2.5 \)
Step 3: Calculate heat rejected using energy conservation: \( Q_H = Q_L + W = 200 + 80 = 280 \) J
Answer
COP of the refrigerator = 2.5 (dimensionless). Heat rejected to surroundings = 280 J.
Example 2 (Class 11-12 Level — Carnot COP)
Problem: A Carnot refrigerator operates between a cold reservoir at 5°C and a hot reservoir at 35°C. Calculate (a) the COP of the refrigerator, (b) the COP of the equivalent heat pump, and (c) the work needed to remove 1500 J of heat from the cold reservoir.
Given: \( T_L = 5 + 273 = 278 \) K, \( T_H = 35 + 273 = 308 \) K, \( Q_L = 1500 \) J
Step 1: Apply the Carnot COP formula for a refrigerator:
\[ COP_{Carnot} = \frac{T_L}{T_H – T_L} = \frac{278}{308 – 278} = \frac{278}{30} \approx 9.27 \]
Step 2: Use the relation \( COP_{HP} = COP_R + 1 \):
\[ COP_{HP} = 9.27 + 1 = 10.27 \]
Step 3: Find the work input using \( COP_R = Q_L / W \):
\[ W = \frac{Q_L}{COP_R} = \frac{1500}{9.27} \approx 161.8 \text{ J} \]
Answer
(a) COP of refrigerator ≈ 9.27. (b) COP of heat pump ≈ 10.27. (c) Work required ≈ 161.8 J.
Example 3 (JEE Level — Concept Application)
Problem: A heat pump is used to heat a room maintained at 20°C. The outside temperature is −10°C. The heat pump delivers 3 kW of heat to the room. Assuming Carnot performance, find (a) the minimum power consumed by the heat pump, and (b) the rate of heat extracted from the outside air.
Given: \( T_H = 20 + 273 = 293 \) K, \( T_L = -10 + 273 = 263 \) K, \( \dot{Q}_H = 3000 \) W
Step 1: Calculate the Carnot COP of the heat pump:
\[ COP_{HP} = \frac{T_H}{T_H – T_L} = \frac{293}{293 – 263} = \frac{293}{30} \approx 9.77 \]
Step 2: Find minimum power (work rate) using \( COP_{HP} = \dot{Q}_H / \dot{W} \):
\[ \dot{W} = \frac{\dot{Q}_H}{COP_{HP}} = \frac{3000}{9.77} \approx 307 \text{ W} \]
Step 3: Find the rate of heat extracted from outside air using energy conservation:
\[ \dot{Q}_L = \dot{Q}_H – \dot{W} = 3000 – 307 = 2693 \text{ W} \approx 2.69 \text{ kW} \]
Answer
(a) Minimum power consumed ≈ 307 W. (b) Rate of heat extracted from outside ≈ 2.69 kW.
CBSE Exam Tips 2025-26
- Always convert temperatures to Kelvin. The most common error in COP problems is using Celsius directly in the Carnot formula. Add 273 (or 273.15 for precision) to every temperature before substituting.
- Memorise the key relation \( COP_{HP} = COP_R + 1 \). This relation is frequently tested in CBSE 3-mark questions. We recommend deriving it once from first principles so you never forget it.
- Know both forms of the COP formula. CBSE questions may give you \( Q_L \), \( Q_H \), and \( W \), or they may give temperatures. Practice switching between \( COP = Q_L/W \) and \( COP = T_L/(T_H – T_L) \).
- State the unit clearly. COP is dimensionless. In 2025-26 board exams, writing “no unit” or “dimensionless” in your answer earns the full mark for units.
- Draw a labelled diagram. A simple block diagram showing the cold reservoir, refrigerator/heat pump, hot reservoir, and arrows for \( Q_L \), \( Q_H \), and \( W \) can earn 1 bonus mark in descriptive questions.
- Understand the physical meaning. A COP greater than 1 is possible and normal for refrigerators. Do not confuse COP with efficiency, which is always less than 1 for a heat engine.
Common Mistakes to Avoid
- Using Celsius instead of Kelvin in the Carnot COP formula. The formula \( COP = T_L / (T_H – T_L) \) requires absolute temperatures. Always convert: \( T(K) = T(°C) + 273 \). Note that the temperature difference \( (T_H – T_L) \) is the same in both Kelvin and Celsius, but the ratio \( T_L / (T_H – T_L) \) is not.
- Confusing COP of a refrigerator with COP of a heat pump. For a refrigerator, the useful output is \( Q_L \) (cooling). For a heat pump, the useful output is \( Q_H \) (heating). Always identify the device type first.
- Forgetting the relation \( W = Q_H – Q_L \). Many students try to find \( W \) without using energy conservation. This relation is always valid for any refrigerator or heat pump cycle.
- Assuming COP must be less than 1. Unlike heat engine efficiency, the COP of a refrigerator or heat pump is commonly greater than 1. A Carnot refrigerator between 0°C and 30°C has a COP of about 9.1, which is perfectly valid.
- Mixing up source and sink temperatures. For a refrigerator, \( T_L \) is the cold interior temperature and \( T_H \) is the room temperature. Swapping them gives a negative or incorrect COP. Always check which reservoir is hotter.
JEE/NEET Application of the COP Formula
In our experience, JEE aspirants encounter the COP formula in the Thermodynamics chapter, which typically carries 2–3 questions in JEE Main and 1–2 questions in NEET. The formula appears in three main patterns.
Pattern 1: Direct COP Calculation
JEE Main often gives \( Q_L \), \( Q_H \), or \( W \) and asks for COP. The key is to correctly identify which formula to apply. If the device is a refrigerator, use \( COP_R = Q_L / W \). If it is a heat pump, use \( COP_{HP} = Q_H / W \). These are straightforward one-step problems worth 4 marks each.
Pattern 2: Carnot COP and Temperature-Based Problems
JEE Advanced frequently combines the Carnot COP formula with the concept of reversibility. A typical problem gives the temperatures of two reservoirs and asks for the minimum work needed to transfer a given amount of heat. The minimum work corresponds to maximum (Carnot) COP. Use \( COP_{Carnot} = T_L / (T_H – T_L) \) and then \( W = Q_L / COP_{Carnot} \). NEET also tests this pattern in single-correct MCQ format.
Pattern 3: Combined Heat Engine and Refrigerator Problems
In advanced JEE problems, a heat engine drives a refrigerator. The work output of the heat engine equals the work input of the refrigerator. You must apply the efficiency formula \( \eta = 1 – T_L/T_H \) to the engine and the COP formula to the refrigerator simultaneously. Our experts suggest practising at least 5 such combined problems before the exam. The relation \( COP_{HP} = COP_R + 1 \) is also tested as a conceptual MCQ in both JEE and NEET.
FAQs on COP Formula
Explore More Physics Formulas
Now that you have mastered the COP Formula, strengthen your Thermodynamics preparation with these related resources on ncertbooks.net. Study the Fluid Mechanics Formula to understand pressure and flow relationships in Class 11 Physics. Review the Angular Velocity Formula for rotational mechanics problems in JEE. Revisit the Normal Force Formula to solidify your understanding of Newton's laws. For the complete collection of Physics formulas aligned with NCERT and CBSE 2025-26 syllabus, visit our Physics Formulas Hub. You can also verify thermodynamics concepts directly from the official NCERT website.