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Heat flows from hot to cold on its own. A refrigerator, an air conditioner and a heat pump all push it the other way, and the question exam papers ask is: how much work does that cost? Students who have learned heat engine efficiency often carry the wrong formula across, or plug in Celsius temperatures. This post sets out the definitions, works through four problems with numbers, and lists the traps.
Where does this topic sit in the syllabus?
The NCERT Class 11 chapter Thermodynamics states that the second law "gives a fundamental limitation to the efficiency of a heat engine and the co-efficient of performance of a refrigerator", and that for a refrigerator "the co-efficient of performance can never be infinite". It gives the Clausius statement: no process is possible whose sole result is the transfer of heat from a colder object to a hotter object. It also explains that running each step of the Carnot cycle in reverse gives a reversible refrigerator.
The Syllabus for JEE (Main) 2026 lists the second law and reversible and irreversible processes, and the JEE (Advanced) 2026 Information Brochure adds "Carnot engine and its efficiency". Neither names refrigerators separately, so treat them as an application of the second law and the Carnot cycle, and check the syllabus for your exam year.
How is the coefficient of performance defined?
Let a device take heat Q₂ from a cold reservoir at temperature T₂, receive work W, and deliver heat Q₁ to a hot reservoir at T₁. Energy conservation over a cycle gives Q₁ = Q₂ + W.
Refrigerator or air conditioner. The useful effect is heat removed from the cold side: β = Q₂ / W.
Heat pump (heating mode). The useful effect is heat delivered to the warm side: COP_HP = Q₁ / W = β + 1.
For a reversible (Carnot) device the heats are proportional to absolute temperatures, Q₁/Q₂ = T₁/T₂, which gives the upper limits
β_max = T₂ / (T₁ − T₂) and COP_HP,max = T₁ / (T₁ − T₂).
Both can be much larger than 1. That is not a free lunch: the device moves heat that already exists rather than creating it from work.
Worked example 1: a kitchen refrigerator
A refrigerator with β = 3.0 removes heat from its interior at 150 W. How much electrical power does it use, and how much heat goes into the kitchen?
W = Q₂ / β = 150 / 3.0 = 50 W.
Heat to kitchen = Q₂ + W = 150 + 50 = 200 W.
This is why the back of a fridge is warm, and why leaving its door open heats the kitchen overall.
Upper limit. With the interior at 4 °C (277 K) and the kitchen at 30 °C (303 K), β_max = 277 / 26 ≈ 10.7. The real fridge achieves less than a third of the ideal, which is typical of irreversible real machines.
Worked example 2: a heat pump warming a room
A heat pump keeps a room at 20 °C (293 K) when it is 0 °C (273 K) outside. It uses 2.0 kW of electrical power and has a COP of 3.5.
Heat delivered: Q₁ = 3.5 × 2.0 = 7.0 kW.
Heat taken from outside: Q₂ = Q₁ − W = 7.0 − 2.0 = 5.0 kW.
Ideal limit: COP_HP,max = 293 / 20 ≈ 14.7.
An electric heater using the same 2.0 kW delivers 2.0 kW of heat. The heat pump delivers more because most of its output is heat pulled in from the cold air.
Worked example 3: making ice
How much work does a freezer with β = 4.0 need to turn 1.0 kg of water at 0 °C into ice at 0 °C? NCERT's Thermal Properties of Matter gives the latent heat of fusion of water as 3.33 × 10⁵ J/kg.
Heat to remove: Q₂ = 1.0 × 3.33 × 10⁵ = 3.33 × 10⁵ J.
Work: W = Q₂ / β = 3.33 × 10⁵ / 4.0 ≈ 8.3 × 10⁴ J.
Heat rejected: Q₁ = 3.33 × 10⁵ + 0.83 × 10⁵ ≈ 4.2 × 10⁵ J.
With a 250 W compressor, the time is 8.3 × 10⁴ / 250 ≈ 333 s, about 5.6 minutes (ignoring heat leaking in).
Worked example 4: why air conditioners struggle on hot days
Keep a room at 25 °C (298 K). If outside is 35 °C (308 K), β_max = 298 / 10 ≈ 29.8. If outside is 45 °C (318 K), β_max = 298 / 20 ≈ 14.9. Doubling the temperature gap halves the ideal coefficient of performance, so the same cooling needs about twice the work in the ideal case.
Why do students find this hard?
Cochran and Heron (2006) asked students in courses from algebra-based introductory physics up to junior-level thermodynamics whether certain processes in heat engines and refrigerators could occur. Their answers revealed several conceptual difficulties, including failing to see that the second law was relevant at all. Research-based tutorials improved performance on exam questions. In practice, this means checking every proposed device against both laws: energy must balance, and the claimed β must not beat the Carnot limit.
Which mistakes are most common?
Calling β an efficiency. Efficiency cannot exceed 1; a coefficient of performance usually does.
Swapping Q₁ and Q₂. For a fridge, the useful heat is the heat removed, Q₂.
Forgetting Q₁ = Q₂ + W. The hot side always receives more heat than was removed.
Using β + 1 for different reservoirs. COP_HP = β + 1 holds for the same machine between the same temperatures.
Accepting impossible claims. If a problem's device beats T₂/(T₁ − T₂), the answer is "not possible".
For the engine side of the same cycle, read our guide to heat engines and the Carnot cycle, and for signs of Q and W, our post on thermodynamics sign conventions.
A practice routine
Draw the three-arrow diagram (Q₂ in, W in, Q₁ out) for every problem.
Solve one problem of each type: fridge, heat pump, ice making, and "is this device possible?".
For each, compute the Carnot limit and compare.
If you are not sure where to begin with thermodynamics as a whole, Fermi's AI tutor can work out what you already know and build a plan for the session, so you start with the right piece.
Check one appliance at home: find its rated cooling and power input, and work out its coefficient of performance.
