In short

A refrigerator is a cyclic device that absorbs heat Q_C from a cold reservoir (the food compartment), receives work input W from an external source (the compressor motor, plugged into the wall), and dumps heat Q_H = Q_C + W to a hot reservoir (the kitchen air, via the condenser coils at the back). Its performance is measured by the coefficient of performance (COP):

\boxed{\;\text{COP}_{\text{ref}} \;=\; \frac{Q_C}{W} \;=\; \frac{Q_C}{Q_H - Q_C}\;}

A heat pump is the same machine used for a different purpose — to warm a space by moving heat from outside to inside. Its COP is defined with Q_H (the useful heat delivered to the warm space) in the numerator:

\boxed{\;\text{COP}_{\text{hp}} \;=\; \frac{Q_H}{W}\;}

The two COPs are linked by a simple identity that follows straight from the first law:

\text{COP}_{\text{hp}} \;=\; \text{COP}_{\text{ref}} + 1.

Both are bounded above by a Carnot limit that depends only on the two reservoir temperatures (in kelvin):

\text{COP}_{\text{ref, Carnot}} \;=\; \frac{T_C}{T_H - T_C}, \qquad \text{COP}_{\text{hp, Carnot}} \;=\; \frac{T_H}{T_H - T_C}.

A Godrej 300 L refrigerator running with food at 4 °C and kitchen air at 30 °C has a Carnot ceiling of COP ≈ 10.7; a real inverter unit achieves about 3 to 4. A Delhi split AC pumping heat out of a 24 °C room into 45 °C afternoon air has Carnot COP ≈ 14, real COP ≈ 3 — and the gap grows as the outside air gets hotter, which is why AC bills explode in May.

Open the door of a Godrej 300-litre single-door refrigerator in a Chennai kitchen at 1 p.m. The food compartment is at about 4 °C. The freezer is at −18 °C. You reach in, pull out a bottle of cold water, and close the door. Now walk around to the back. Put your hand on the black wire-grille on the outer wall. It is warm to the touch — 40 to 45 °C, hotter than the room. Behind that grille, inside the compressor chamber, a small motor is buzzing.

Something strange is happening. The room is at 30 °C. The food inside is at 4 °C. Heat, if left alone, flows from hot to cold — that is the Clausius statement of the second law. But inside that fridge, heat is somehow flowing out of the cold food compartment and into the already-hotter kitchen. The direction is wrong. Something is pushing heat up a temperature hill.

That something is the compressor. It is drawing electrical energy from the 230 V mains, converting it to mechanical work on a gas (the refrigerant), and using that work to ratchet heat out of the cold space and into the hot one. A fridge is not a "cold generator"; it is a heat pump running in reverse, and the work you pay for — on your monthly electricity bill — is the cost of violating the natural direction of heat flow.

This article is the full thermodynamic construction. Start from the first law applied to a cyclic device; define the coefficient of performance; prove the refrigerator-heat-pump identity \text{COP}_{\text{hp}} = \text{COP}_{\text{ref}} + 1; derive the Carnot bounds by running the Carnot cycle backwards; explain how a real Voltas vapour-compression refrigeration cycle works (evaporator, compressor, condenser, expansion valve — the four organs of every fridge, AC, and cold-chain truck in the country); walk through three worked examples grounded in Indian kitchens and Delhi summers; and close with two big questions — how efficient can a fridge be, and why do heat pumps beat resistive heaters by a factor of three for home heating, even though a resistive heater is "100 % efficient."

A refrigerator is a reversed heat engine — the universal schematic

A heat engine takes heat from a hot reservoir, performs work, and rejects waste heat to a cold reservoir. Reverse every arrow in that diagram and you get a refrigerator: heat flows from the cold reservoir into the device, work flows from outside into the device, and heat flows from the device out to the hot reservoir. The device is cyclic — its working substance (a refrigerant) returns to its starting state every cycle — so the internal energy of the device itself is unchanged over a full cycle: \Delta U = 0.

The first law applied over one full cycle gives

\Delta U \;=\; Q_{\text{net}} \;-\; W_{\text{net}} \;=\; 0.

For a refrigerator, reading signs as positive magnitudes of the arrows in the schematic:

Heat gained by the device is Q_C; heat lost by the device is Q_H; work gained by the device is W. Over the cycle the device returns to its starting state, so energy-in equals energy-out:

Q_C \;+\; W \;=\; Q_H. \tag{1}

Why: the device cannot store energy across a cycle (state function, \Delta U = 0). Every joule that came in as heat or work must leave. Two things came in — Q_C and W — and one thing went out, Q_H. Their magnitudes must balance.

This equation will do nearly all the book-keeping in this article. It is worth staring at. It says the heat dumped at the back of the fridge is always more than the heat extracted from the inside, by exactly the amount of work the compressor did. The warm air you feel at the back of a Godrej fridge contains, in joules, all the heat pulled out of your vegetables plus the electricity you paid for. The fridge heats your kitchen every time it cools your food — a fact you can verify at 2 a.m. with a thermometer in front of the condenser grille.

Universal schematic of a refrigerator A hot reservoir at the top labelled T_H equals kitchen at thirty Celsius; a cold reservoir at the bottom labelled T_C equals food at four Celsius; a cyclic device in the middle labelled compressor cycle. Heat Q_C flows from cold reservoir up into the device; work W flows horizontally into the device from the right; heat Q_H flows up and out of the device into the hot reservoir. The arrow Q_H equals Q_C plus W by energy conservation. hot reservoir — kitchen at T_H ≈ 303 K (30 °C) cold reservoir — food at T_C ≈ 277 K (4 °C) compressor cycle (ΔU=0) Q_C + W = Q_H Q_C Q_H W (compressor motor)
The universal refrigerator schematic. Energy conservation across one cycle forces $Q_H = Q_C + W$: the heat dumped to the kitchen is the heat extracted from the food plus the electrical work paid to the compressor.

The coefficient of performance

For a heat engine, "efficiency" means what you get (work) divided by what you pay for (heat). For a refrigerator, the logic is the same but the quantities are swapped. You pay for the electrical work W you feed into the compressor. What you want is the heat Q_C pulled out of the cold space — that is what keeps your dal safe overnight. The natural figure of merit is

\boxed{\;\text{COP}_{\text{ref}} \;=\; \frac{Q_C}{W}\;}. \tag{2}

Why: same "benefit over cost" structure as engine efficiency, but with Q_C (the cooling delivered) as the benefit and W (the electrical work purchased) as the cost. The name coefficient of performance — rather than efficiency — is used because the ratio can exceed 1: you can pull 3 J of heat out of the fridge for every 1 J of electrical work, which would sound absurd if you called it "efficiency."

A COP of 1 means every joule of electricity buys one joule of cooling. A COP of 3 means every joule of electricity buys three joules of cooling. Commercial fridges typically achieve COP in the range 2 to 4, so every rupee you spend on electricity pays for 2 to 4 rupees of cooling (valued in thermal terms).

Using Q_H = Q_C + W to eliminate W:

W = Q_H - Q_C \;\Rightarrow\; \text{COP}_{\text{ref}} \;=\; \frac{Q_C}{Q_H - Q_C}. \tag{2'}

Why: substitute directly from the first-law identity. This form is useful when you know the heat flows on both sides and want to back out the COP without measuring the work separately.

A heat pump — same device, different emphasis

A heat pump is exactly the same hardware as a refrigerator — a compressor, evaporator, condenser, and expansion valve — but installed to heat a space rather than cool one. Instead of measuring the heat pulled from the cold side (what you want from a fridge), you measure the heat delivered to the hot side (what you want from a heater).

A winter heat-pump installation in Shimla might have the cold side outside (−5 °C outdoor air) and the hot side inside (22 °C living room). The pump draws heat from the cold outdoor air, does work on it, and delivers the sum to the indoor air. Its figure of merit is

\boxed{\;\text{COP}_{\text{hp}} \;=\; \frac{Q_H}{W}\;}. \tag{3}

Why: the "benefit" has shifted from "heat removed from inside" (Q_C) to "heat delivered to inside" (Q_H). The denominator — work purchased from the mains — is unchanged.

A domestic heat pump used for winter heating in Shimla, or a commercial geothermal heat pump at an IT park in Pune, aims for COP around 3 to 5 — i.e., it delivers 3 to 5 joules of heat indoors for every joule of electricity paid. This is the killer economic argument for heat pumps: a resistive electric heater has a "COP" of exactly 1 (every joule of electricity becomes one joule of heat), while a heat pump routinely triples or quadruples that. You do not get energy for free — the extra heat was always out there in the cold air; the pump just moves it uphill.

The refrigerator-heat-pump identity

Divide equation (1) by W:

\frac{Q_C}{W} + 1 \;=\; \frac{Q_H}{W},

which gives the exact identity

\boxed{\;\text{COP}_{\text{hp}} \;=\; \text{COP}_{\text{ref}} + 1\;}. \tag{4}

Why: the heat-pump COP is the refrigerator COP plus one, because the heat delivered to the hot side (Q_H) always exceeds the heat extracted from the cold side (Q_C) by exactly the work W — and dividing that "extra W" by W gives exactly 1. A Godrej fridge with COP 3 is mathematically the same as a heat pump with COP 4 running between the same two reservoirs.

This is not an approximation. It is energy conservation, rewritten. Any device that claims COP_ref + 1 ≠ COP_hp is violating the first law.

The Carnot bound — best possible COP

You have now seen the definition of COP. Nothing has said anything about its maximum value — why can't you make a refrigerator with COP = 1000? The answer comes from running the Carnot cycle in reverse, and using the second law to show that no refrigerator can beat it.

Reversed Carnot: the same four legs, run backwards

A Carnot heat engine runs:

  1. Isothermal expansion at T_H — absorbs Q_H.
  2. Adiabatic expansion — cools from T_H to T_C, no heat exchange.
  3. Isothermal compression at T_C — rejects Q_C.
  4. Adiabatic compression — heats from T_C to T_H, no heat exchange.

Net work output: W = Q_H - Q_C. Efficiency: \eta = 1 - T_C/T_H.

Run this cycle in reverse (every arrow flipped), and you get a Carnot refrigerator:

  1. Adiabatic expansion — the refrigerant cools from T_H to T_C, no heat flow.
  2. Isothermal expansion at T_C — refrigerant absorbs heat Q_C from the cold space.
  3. Adiabatic compression — refrigerant heats from T_C to T_H, no heat flow.
  4. Isothermal compression at T_H — refrigerant rejects heat Q_H to the hot space.

Net work input: W = Q_H - Q_C. The heat magnitudes Q_H and Q_C are the same as in the forward cycle, because the cycle is reversible — reversing the path reverses every heat flow's sign but keeps every magnitude.

For the forward Carnot cycle, the heats satisfy (derived in heat engines and carnot cycle)

\frac{Q_C}{Q_H} \;=\; \frac{T_C}{T_H}. \tag{5}

This is a clean consequence of computing Q for each isothermal leg using the ideal gas law (Q = nRT \ln(V_f/V_i)) and noting that the adiabatic legs force V-ratios to match.

Carnot COP — the formula

Apply equation (5) to the refrigerator COP:

\text{COP}_{\text{ref, Carnot}} \;=\; \frac{Q_C}{Q_H - Q_C} \;=\; \frac{Q_C/Q_H}{1 - Q_C/Q_H} \;=\; \frac{T_C/T_H}{1 - T_C/T_H}.

Multiply numerator and denominator by T_H:

\boxed{\;\text{COP}_{\text{ref, Carnot}} \;=\; \frac{T_C}{T_H - T_C}\;} \tag{6}

Why: start with the definition of COP in heat form, substitute the Carnot-cycle relation Q_C/Q_H = T_C/T_H, and simplify. The result depends only on the two reservoir temperatures — not on the refrigerant (freon, ammonia, propane, isobutane, CO₂), not on the pressure range, not on the compressor design.

Apply equation (5) to the heat-pump COP:

\text{COP}_{\text{hp, Carnot}} \;=\; \frac{Q_H}{Q_H - Q_C} \;=\; \frac{1}{1 - Q_C/Q_H} \;=\; \frac{1}{1 - T_C/T_H} \;=\; \frac{T_H}{T_H - T_C}.
\boxed{\;\text{COP}_{\text{hp, Carnot}} \;=\; \frac{T_H}{T_H - T_C}\;} \tag{7}

You can check the identity (4) explicitly:

\text{COP}_{\text{hp, Carnot}} - \text{COP}_{\text{ref, Carnot}} \;=\; \frac{T_H - T_C}{T_H - T_C} \;=\; 1. \;\checkmark

Why Carnot is the upper bound — in one clean argument

Claim: no refrigerator operating between T_H and T_C can have a COP greater than the Carnot COP between those temperatures.

Proof by contradiction. Suppose a fridge F achieves \text{COP}_F > \text{COP}_{\text{Carnot}} between the same two reservoirs. That means F extracts more heat Q_C from the cold reservoir per unit of work W than the Carnot fridge C does. Now build a composite:

Now compute the net heat flow between the two reservoirs. By assumption, F has a higher COP than the Carnot (engine-reversed) fridge, which means it delivers more cold-side extraction per joule of work. Using W_E = W from the engine to drive F, the composite (E + F) has zero net work exchanged with the outside world and transfers net heat from the cold reservoir to the hot reservoir with no external work. That violates the Clausius statement of the second law — heat does not spontaneously flow from cold to hot.

Therefore the assumption \text{COP}_F > \text{COP}_{\text{Carnot}} is false. No refrigerator beats the Carnot COP. The same argument applied with the engine and fridge swapped proves that equality is achievable only by reversible (Carnot) fridges; real irreversibilities always push COP down.

This is exactly Carnot's theorem for refrigerators — and it is the fridge version of the same argument that bounds engine efficiency. The second law caps both machines from the same direction.

Numbers: what Carnot lets us do in Indian conditions

The Carnot COP depends only on temperatures in kelvin. Some cases:

Scenario T_C (K) T_H (K) T_H - T_C (K) Carnot COP_ref
Godrej kitchen fridge 277 (4 °C) 303 (30 °C) 26 10.7
Godrej kitchen fridge in summer 277 (4 °C) 318 (45 °C) 41 6.75
Freezer compartment 255 (−18 °C) 303 (30 °C) 48 5.31
Split AC Delhi, mild 297 (24 °C) 313 (40 °C) 16 18.6
Split AC Delhi, peak May 297 (24 °C) 323 (50 °C) 26 11.4
Cold-chain truck (vaccines) 275 (2 °C) 318 (45 °C) 43 6.40
Liquid nitrogen cold finger 77 (−196 °C) 300 (27 °C) 223 0.35

Two big lessons jump off this table.

Lesson 1: COP rises steeply as the temperature gap T_H - T_C shrinks. A fridge pumping heat across 26 K can have COP 10.7; across 48 K (the freezer) only 5.3. Cold storage for vaccines, which needs to maintain 2 °C even when outside is 45 °C, faces a 43-K gap with Carnot ceiling 6.4 — three times worse than a domestic fridge. This is why cold-chain logistics across the Indian subcontinent in May are expensive: you are operating at a thermodynamic disadvantage compared to winter.

Lesson 2: summer is brutal. A Delhi split AC that achieves a Carnot COP of 18.6 in mild weather (40 °C outside) drops to 11.4 when the outside hits 50 °C. In practice, real ACs have COP around 3 to 4 — already far below Carnot — but the trend is the same: when outside temperature rises, COP falls, and your electricity bill climbs faster than linearly with temperature.

Explore: drag the temperature difference

The Carnot COP depends almost entirely on the temperature gap T_H - T_C (for fixed T_C, it falls as T_C/\Delta T). Drag the red dot below to change the outside temperature T_H from 20 °C to 55 °C while keeping the inside at T_C = 4 °C. Watch the Carnot COP drop precipitously as outside gets hotter.

Interactive: Carnot COP as a function of outside temperature A falling curve of Carnot refrigerator COP plotted against outside temperature. A draggable red dot moves along the outside temperature axis with a readout showing the current COP. Inside temperature is fixed at 4 Celsius. outside air temperature T_H (°C), with T_C = 4 °C Carnot COP_ref 5 10 15 20 25 30 35 40 45 50 55 drag the red point along the axis
Carnot-bounded COP for a fridge keeping the inside at 4 °C, as a function of outside temperature. The COP falls as a $1/\Delta T$ hyperbola. Every 10 K the outside heats up, the Carnot ceiling drops by a factor of roughly 1.3 — and since real COP sits below the ceiling, the real device suffers too. This is the quantitative skeleton of the Delhi summer electricity bill.

The vapour-compression cycle — what is inside a Godrej fridge

The Carnot cycle is an idealisation. Real refrigerators use a practical cousin: the vapour-compression cycle, which exploits the latent heat of vaporisation (see phase transitions and latent heat) of a refrigerant fluid to move large amounts of heat with compact hardware. Every domestic fridge, split AC, car AC, and cold-chain truck in India runs a variant of this cycle. The four components:

  1. Evaporator (inside the cold space, behind the back wall of the fridge compartment). Low-pressure liquid refrigerant enters; it is below its boiling point at this pressure, and it absorbs Q_C from the food compartment, vaporising into a low-pressure gas. The heat absorbed is Q_C = m L_v for the mass m of refrigerant that vaporised.
  2. Compressor (the buzzing motor at the back). Low-pressure gas is compressed to high pressure. The gas heats up in the process because compression is close to adiabatic. The work done on the gas is W; electricity from the mains flows into the motor and out of the motor's shaft as mechanical work.
  3. Condenser (the black wire grille on the outer back of the fridge). High-pressure hot gas enters. Because it is now hotter than the kitchen air, it dumps heat Q_H out through the grille and condenses back into liquid. The heat released is Q_H = m L_v' for the same mass m, at the higher pressure and slightly different temperature.
  4. Expansion valve (or capillary tube). High-pressure liquid is throttled (Joule-Thomson expansion) back to low pressure; this cools it rapidly. The cold, low-pressure liquid now enters the evaporator to start the cycle over.

The refrigerant — historically CFCs (Freon-12), then HCFCs (R-22), now HFCs (R-134a, R-410A), and increasingly natural refrigerants like propane (R-290) in Indian five-star rated fridges — is chosen so that at the evaporator pressure it boils at a temperature below the food compartment (so heat flows in), and at the condenser pressure it condenses at a temperature above the kitchen (so heat flows out). The entire trick of refrigeration is the tuning of the refrigerant's phase-transition temperatures with pressure. That is why phase transitions and this chapter are so tightly linked.

Schematic of a vapour-compression refrigeration cycle Four components arranged in a loop: an evaporator at the lower left inside the cold food compartment; a compressor at the lower right driven by mains electricity; a condenser at the upper right in the warm kitchen; an expansion valve at the upper left. Refrigerant flows in the loop: vapour from evaporator to compressor, hot compressed vapour from compressor to condenser, liquid from condenser to expansion valve, low-pressure mixture from expansion valve to evaporator. Q_C enters the evaporator from the food, Q_H leaves the condenser to the kitchen, W enters the compressor from the mains. kitchen side: hot reservoir T_H food compartment: cold reservoir T_C condenser (hot gas → liquid) Q_H expansion valve (throttle, cools) evaporator (liquid → cold gas) Q_C compressor (pumps gas) W (mains) cold vapour, low pressure hot vapour, high pressure hot liquid, high pressure cold liquid, low pressure
The four organs of every domestic refrigerator and AC: evaporator, compressor, condenser, expansion valve. Refrigerant cycles around the loop — absorbing heat $Q_C$ at the cold end (boiling in the evaporator) and releasing heat $Q_H$ at the warm end (condensing in the condenser). The compressor drives the cycle, consuming work $W$ from the mains.

Watch the refrigerant travel

The animation below shows a refrigerant "packet" moving around the four-component loop in real time. At each stage, its phase (liquid or vapour) and its position in the cycle are indicated. Use it to see the cycle as a closed loop rather than a list of stations.

Animated refrigerant cycle A dot representing a refrigerant packet travels clockwise around a rectangular loop. It moves left-to-right along the bottom (cold vapour), upward on the right (compressed hot vapour), right-to-left along the top (hot liquid), and downward on the left (cold liquid, post-expansion). condenser expansion valve evaporator compressor hot side — kitchen air T_H cold side — food compartment T_C
A single refrigerant packet traced around the vapour-compression loop. Along the bottom (evaporator → compressor): cold, low-pressure vapour. Up the right (compressor): compression, heating, rising to high pressure. Along the top (condenser → expansion valve): hot, high-pressure liquid releasing $Q_H$ to the kitchen. Down the left (expansion → evaporator): cold, low-pressure liquid, ready to absorb $Q_C$ again. Click replay to watch another cycle.

Worked examples

Example 1: Godrej fridge on a Chennai afternoon

A Godrej 300 L single-door refrigerator is running in a Chennai kitchen at 1 p.m. The interior (food compartment) is held at 4 °C. The kitchen air is at 32 °C. The fridge's compressor draws 120 W from the mains on average, and it has a measured cooling capacity (heat extraction rate) of 420 W from the food compartment. (a) What is its real COP? (b) What is the Carnot COP between these two reservoirs? (c) What fraction of the ideal is the real unit achieving?

Energy balance diagram for the Godrej fridge worked example A fridge cabinet with Q_C equal to 420 watts arriving from the food into the device, W equal to 120 watts from the mains, and Q_H equal to 540 watts being dumped out to the kitchen, showing the first-law balance Q_H equals Q_C plus W. Godrej 300 L interior 4 °C Q_C = 420 W W = 120 W Q_H = 540 W
The fridge extracts 420 W from the food and draws 120 W from the wall; by first-law book-keeping it dumps 540 W to the kitchen air at the back.

Step 1. Compute the real COP.

\text{COP}_{\text{real}} \;=\; \frac{Q_C}{W} \;=\; \frac{420}{120} \;=\; 3.50.

Why: definition of COP for a refrigerator: cooling delivered per unit work purchased. Q_C here is a rate (W = J/s), and so is W, but the ratio is dimensionless — a COP.

Step 2. Compute the Carnot COP between T_C = 277.15 K and T_H = 305.15 K.

\text{COP}_{\text{Carnot}} \;=\; \frac{T_C}{T_H - T_C} \;=\; \frac{277.15}{305.15 - 277.15} \;=\; \frac{277.15}{28} \;=\; 9.90.

Why: Carnot's formula depends only on the two reservoir temperatures in kelvin. Converting 4 °C and 32 °C to kelvin gives 277.15 K and 305.15 K. Their difference is 28 K, and the Carnot COP is the cold-side temperature divided by that difference.

Step 3. Compute the fraction of Carnot achieved.

\frac{\text{COP}_{\text{real}}}{\text{COP}_{\text{Carnot}}} \;=\; \frac{3.50}{9.90} \;=\; 0.354 \;\approx\; 35.4\%.

Why: real refrigerators suffer irreversibilities — friction in the compressor, heat leaks through the insulation, finite-rate heat transfer across the coils — that push the COP below its Carnot ceiling. Reaching 35 % of Carnot is typical for a modern five-star-rated domestic fridge; older units can be as low as 20 %.

Step 4. Check the heat-pump identity.

Q_H \;=\; Q_C + W \;=\; 420 + 120 \;=\; 540\,\text{W},
\text{COP}_{\text{hp}} \;=\; Q_H/W \;=\; 540/120 \;=\; 4.50 \;=\; 3.50 + 1. \;\checkmark

Result: Real COP_ref = 3.50, Carnot COP_ref = 9.90, achievement = 35 %. The fridge also dumps 540 W of heat to the kitchen continuously — enough to warm a tightly sealed kitchen noticeably over the course of a hot afternoon.

What this shows: The fridge converts every joule of electricity into 3.5 J of cooling, but also (by first-law book-keeping) into 4.5 J of heat released to the kitchen. Running a fridge in an already-warm, poorly-ventilated kitchen is a losing game: the heat it dumps can partially negate the cooling of the AC in the same room. This is why commercial restaurants pipe their refrigerator condenser heat outside the building.

Example 2: Voltas split AC in Delhi in May

A Voltas 1.5-ton split AC is installed in a Delhi bedroom. "1.5 ton" in AC jargon means a rated cooling capacity of 1.5 \times 3517 = 5276 W ≈ 5.3 kW (one ton of refrigeration is 3517 W, historically the rate at which one ton of ice melts in 24 hours). The AC maintains the room at 24 °C while the outside air is at 45 °C. The manufacturer specifies an ISEER (inverter seasonal energy efficiency ratio) of 4.2, which for this calculation we will treat as the COP in operation. (a) What compressor electrical power does it consume at full rated cooling? (b) What is the Carnot ceiling at these temperatures? (c) What is the efficiency deficit — and what happens to it if outside climbs to 50 °C (inside still 24 °C)?

Split AC energy flow An indoor unit in a twenty-four-degree room extracting 5.3 kilowatts, connected to an outdoor unit in forty-five-degree air that dumps the condensing heat. Between them, a compressor draws 1.26 kilowatts from the mains. indoor unit (evaporator) room at 24 °C extracts Q_C = 5.3 kW outdoor unit (compressor + condenser) air at 45 °C dumps Q_H = 6.56 kW refrigerant lines W = 1.26 kW
The split AC has its evaporator indoors (pulling heat out of the 24 °C room) and its compressor-plus-condenser outdoors (dumping heat to the 45 °C air). Electrical work enters the outdoor compressor.

Step 1. Compute the electrical power at full rated cooling.

W \;=\; \frac{Q_C}{\text{COP}} \;=\; \frac{5276}{4.2} \;=\; 1256\,\text{W} \;\approx\; 1.26\,\text{kW}.

Why: if the AC is delivering 5.3 kW of cooling at a COP of 4.2, it consumes 5.3/4.2 kW of electricity to do so. This is the approximate compressor draw at full load.

Step 2. Compute the Carnot ceiling at (24 °C indoors, 45 °C outdoors).

T_C \;=\; 297.15\,\text{K}, \quad T_H \;=\; 318.15\,\text{K}, \quad T_H - T_C \;=\; 21\,\text{K}
\text{COP}_{\text{Carnot}} \;=\; \frac{297.15}{21} \;=\; 14.15.

Step 3. Fraction of Carnot at 45 °C.

\frac{4.2}{14.15} \;=\; 0.297 \;\approx\; 29.7\%.

Why: around 30 % of Carnot for a mid-range inverter AC. The difference from the Carnot ideal comes from finite-rate heat transfer across the evaporator and condenser coils, fan power, and compressor inefficiencies.

Step 4. Recompute at T_H = 50 °C (323.15 K) outdoor air — the May peak.

\text{COP}_{\text{Carnot, 50°C}} \;=\; \frac{297.15}{323.15 - 297.15} \;=\; \frac{297.15}{26} \;=\; 11.43.

If the AC still achieves 30 % of Carnot (a reasonable assumption; the same irreversibility mechanisms operate), its real COP at 50 °C drops to

\text{COP}_{\text{real, 50°C}} \;\approx\; 0.30 \times 11.43 \;=\; 3.43.

The compressor power rises to

W_{\text{50°C}} \;=\; \frac{5276}{3.43} \;=\; 1538\,\text{W}.

Why: the Carnot ceiling falls as 1/\Delta T, so the outside getting 5 K hotter shrinks the ceiling from 14.15 to 11.43 (a 19 % drop). Real COP, proportional to the ceiling at fixed mechanical design, drops by the same 19 %. To deliver the same cooling, the compressor must draw 19 % more power — exactly what you observe on your electricity bill when May's heat wave arrives in Delhi.

Result: At 45 °C outside, the 1.5-ton AC draws 1.26 kW; at 50 °C, 1.54 kW. Over a 10-hour afternoon, that is 12.6 kWh versus 15.4 kWh — a 22 % increase in consumption for a 5 K change in outside temperature.

What this shows: The cost of AC in Indian summers is dominated by the outside-temperature dependence of the Carnot ceiling. This is the thermodynamic reason that BEE star ratings depend on a "seasonal" performance index (ISEER) rather than a peak rating — the COP itself varies across the season, and any single number misrepresents reality. It is also why rooftop heat-reflective paint, window shading, and insulation pay back so quickly: lowering the effective T_H at the condenser is mathematically equivalent to buying a more expensive, higher-COP compressor.

Example 3: Heat pump vs resistive heater in Shimla

In a Shimla winter, an inverter-based split AC running in "heat mode" (i.e., as a heat pump) maintains a living room at 22 °C while the outside air is at −2 °C. The indoor unit delivers Q_H = 3.5 kW of heating. Compare the electricity cost of this to a simple resistive (oil-filled radiator) heater that delivers the same 3.5 kW of heat, over an 8-hour evening. Assume the heat pump achieves 34 % of its Carnot COP — typical of a mid-range unit with some frost on the outdoor coil.

Comparison: heat pump vs resistive heater Two panels side by side. Left: heat pump diagram with outside at minus two Celsius, inside at twenty-two Celsius, Q_H equal to 3.5 kilowatts, Q_C from outside air, W from mains computed at 836 watts. Right: resistive heater with W equal to 3.5 kilowatts from mains directly becoming Q_H equal to 3.5 kilowatts, no heat from outside. heat pump COP_hp ≈ 4.19 W = 836 W Q_H = 3500 W (Q_C = 2664 W from outdoor air) resistive heater COP = 1.00 W = 3500 W Q_H = 3500 W (no heat from outside)
Same indoor heating delivered, vastly different electricity draw. The heat pump uses 4.2 times less energy because most of its output came from the outdoor air, not the mains.

Step 1. Compute the Carnot heat-pump COP.

T_C \;=\; 271.15\,\text{K}, \quad T_H \;=\; 295.15\,\text{K}, \quad T_H - T_C \;=\; 24\,\text{K}
\text{COP}_{\text{hp, Carnot}} \;=\; \frac{T_H}{T_H - T_C} \;=\; \frac{295.15}{24} \;=\; 12.30.

Why: in heat-pump mode, the relevant formula is T_H / \Delta T rather than T_C / \Delta T. At 24 K temperature gap, the Carnot ceiling is a lofty 12.3.

Step 2. Note that real machines sit well below Carnot; the exact fraction is taken up in Step 3.

Real heat pumps achieve 30–45 % of their Carnot ceiling, depending on the compressor, coil sizing, and how frost build-up on the outdoor coil is managed. The Carnot ceiling is the theoretical maximum; the manufacturer quotes a real COP that depends on the design.

Step 3. Compute the electrical power for the heat pump.

Modern Indian inverter units used as heat pumps in mild winter conditions typically achieve around 34 % of Carnot. Using that figure:

\text{COP}_{\text{hp, real}} \;=\; 0.34 \times 12.30 \;=\; 4.18.
W_{\text{hp}} \;=\; \frac{Q_H}{\text{COP}_{\text{hp, real}}} \;=\; \frac{3500}{4.18} \;=\; 837\,\text{W}.

Why: fraction of Carnot for a heat pump depends on the quality of the compressor, the coil area, fan speeds, and outdoor air humidity (ice can form on the evaporator in cold weather, spoiling heat transfer). 34 % is a conservative estimate for a mid-range Indian unit; top-tier units achieve 40–45 %.

Step 4. Compute the electrical power for a resistive heater.

W_{\text{resistor}} \;=\; Q_H \;=\; 3500\,\text{W}.

Why: a resistive heater converts electricity to heat with no thermodynamic advantage — every joule in becomes one joule out. Its "COP" is exactly 1 by definition. This is why resistive heating, although cheap to buy, is expensive to run.

Step 5. Compute the 8-hour energy consumption of each, and the cost in rupees at ₹8 per kWh.

\text{heat pump: } 0.836\,\text{kW} \times 8\,\text{h} \;=\; 6.69\,\text{kWh} \;\Rightarrow\; \text{₹}53.5
\text{resistor: } 3.5\,\text{kW} \times 8\,\text{h} \;=\; 28.0\,\text{kWh} \;\Rightarrow\; \text{₹}224

Step 6. Compute the ratio.

\frac{\text{heat pump cost}}{\text{resistor cost}} \;=\; \frac{53.5}{224} \;=\; 0.239 \;\approx\; 24\%.

Result: Heat pump uses ~6.7 kWh and costs ~₹54 for an 8-hour evening's heating. Resistor uses 28 kWh and costs ~₹224. Heat pump saves ₹170 per evening — more than its "extra" installation cost over a season.

What this shows: A heat pump is not a more-efficient heater; it is a differently-organised heater that cheats the second law's apparent prohibition by harvesting heat from outside air. The 2664 W of heat that the outdoor unit pulls from the −2 °C Shimla air is genuine, measurable heat — the outdoor unit's evaporator visibly frosts over in cold weather because of it. Heat pumps are the thermodynamically correct way to heat a building. The economic argument is unanswerable: 4× the effective output per rupee of electricity.

Common confusions

If you came here to understand refrigerators, use the COP formulas, and make sense of your electricity bill, you have what you need. What follows is for readers who want the reverse-Carnot cycle drawn on a PV diagram, the entropy-balance proof that equation (6) really is an upper bound, and the connection to the ozone-layer story that shaped the modern refrigerant industry.

The reverse Carnot cycle on a PV diagram

The forward Carnot cycle — already derived in heat engines and carnot cycle — traces a closed loop on a pressure-volume (PV) diagram, going clockwise: isothermal expansion at T_H, adiabatic expansion to T_C, isothermal compression at T_C, adiabatic compression back to T_H. The enclosed area is the net work output.

Run exactly the same four processes in reverse order, and the cycle goes anticlockwise on the PV diagram. The enclosed area now represents net work input. The heat flows reverse too: instead of absorbing Q_H at T_H (as in the engine), the working substance releases Q_H there; instead of releasing Q_C at T_C, it absorbs Q_C there. The machine is now a refrigerator (or heat pump) rather than an engine.

Because the four processes are the same — two isothermals, two adiabatics — the heat magnitudes on the two isothermal legs satisfy the same relation as in the forward cycle:

\frac{|Q_C|}{|Q_H|} \;=\; \frac{T_C}{T_H}, \tag{8}

derived from Q = nRT\ln(V_f/V_i) on each isothermal and the adiabatic constraint T V^{\gamma - 1} = \text{const} on the two adiabatics (which forces the volume ratios on the two isothermals to be equal).

Substituting (8) into the definition of COP gives equations (6) and (7) exactly, with no further assumptions. This is the clean PV-diagram derivation of the Carnot COP.

The entropy-balance argument

A cleaner, more modern way to derive equation (6) uses entropy (see entropy introduction). For a reversible cycle, the entropy change of the working substance over one full cycle is zero:

\oint \frac{dQ_{\text{rev}}}{T} \;=\; 0.

For the Carnot refrigerator, the two adiabatic legs contribute nothing to this integral (no heat exchanged). The two isothermal legs contribute:

\frac{-Q_H}{T_H} \;+\; \frac{Q_C}{T_C} \;=\; 0 \;\Rightarrow\; \frac{Q_C}{Q_H} \;=\; \frac{T_C}{T_H}.

Why: the isothermal at T_H is a compression (for a refrigerator) during which heat Q_H is rejected by the working substance, so from the substance's point of view the heat flow is negative, giving -Q_H/T_H. The isothermal at T_C is an expansion absorbing heat Q_C, giving +Q_C/T_C. For the cycle entropy change to be zero (reversibility), these must balance.

For a real (irreversible) refrigerator, the Clausius inequality gives \oint dQ/T \leq 0, which translates to Q_C/Q_H < T_C/T_H — and therefore \text{COP} < T_C/(T_H - T_C). This is the clean, modern form of Carnot's theorem for refrigerators. The Carnot COP is the upper bound; real machines fall short by the entropy generated in their irreversibilities.

Refrigerants and the ozone story

The thermodynamics of a refrigerator — the existence of a maximum COP, its T_C/(T_H-T_C) formula, the role of latent heat — is independent of the refrigerant used. But the choice of refrigerant determines everything else: the pressures at which the compressor and condenser must operate, the physical size of the coils, the compressor power required, the noise, the toxicity, the flammability, the environmental impact.

For most of the 20th century, the dominant refrigerants were chlorofluorocarbons (CFCs) such as Freon-12 (dichlorodifluoromethane, CCl₂F₂), developed in the 1930s as a non-toxic, non-flammable, chemically stable alternative to earlier refrigerants (ammonia, sulphur dioxide) which were dangerous. CFCs were a triumph of chemistry — they solved every immediate engineering problem. But they created a new one that took fifty years to notice.

When CFCs leak (and every refrigerator's refrigerant eventually leaks) they rise through the atmosphere largely unreacted — they are chemically stable, which is why they are safe for kitchens. In the stratosphere, ultraviolet radiation breaks their C-Cl bonds, releasing chlorine radicals. Each chlorine radical then catalytically destroys ozone (O₃) — one chlorine radical can destroy tens of thousands of ozone molecules before it is eventually sequestered. The ozone layer thins, particularly over Antarctica where cold stratospheric clouds accelerate the chemistry; the Antarctic ozone hole, first measured in 1985, is the result.

The 1987 Montreal Protocol — one of the most successful international environmental agreements in history — phased CFCs out of production globally. India, which had a substantial domestic CFC industry, complied on a 2010 deadline (developing countries had a 10-year grace period). Modern Indian refrigerators use HFCs (hydrofluorocarbons like R-134a, no chlorine, so no ozone depletion) or increasingly natural refrigerants like R-290 (propane, used in many Godrej and LG 5-star rated units — hydrocarbon, extremely low global warming potential, flammable in large quantities but safe in the small charges used in domestic fridges). The 2016 Kigali Amendment phases down HFCs too because of their high global-warming potential (a kilogram of R-134a in the atmosphere warms the Earth as much as 1430 kg of CO₂).

The story is not a distraction from the physics. It is a consequence of it. The fridge works because the refrigerant's phase-transition temperatures track pressure — but that same chemical stability, which makes the working fluid robust and safe, is what lets it survive long enough in the atmosphere to reach the stratosphere. Every engineering virtue has an environmental cost; the COP formula does not tell you that, but the molecule does.

Where this leads next