THERMODYNAMICS • POWER AND REFRIGERATION CYCLES

COP for Refrigerators & Heat Pumps — Compute coefficient of performance (COP) for refrigerators and heat pumps

Quantify the efficiency of devices that move heat against its natural direction of flow.

Historical Context & Motivation

The quest to move heat from a cold region to a warm one — effectively reversing the natural direction of heat flow — dates back to the early nineteenth century when engineers and physicists first wrestled with the theoretical limits of thermal machines. While early efforts focused on heat engines and the work they could produce, a parallel question arose: how effectively can a device remove heat from a space, or deliver heat to one? The coefficient of performance (COP) emerged as the standard metric to answer that question, providing a dimensionless figure of merit that compares the desired thermal effect to the work input required to achieve it.

1824
Carnot's Theoretical Foundation
Sadi Carnot publishes Réflexions sur la puissance motrice du feu, establishing the maximum efficiency of heat engines and implicitly defining the upper bound for reversed cycles.
1850–1854
Clausius and Kelvin Formalize the Second Law
Rudolf Clausius and Lord Kelvin independently state the Second Law of Thermodynamics, clarifying that heat cannot spontaneously flow from cold to hot and that work input is always required for refrigeration.
1876
Linde's Ammonia Compressor
Carl von Linde patents a practical ammonia-compression refrigeration system, making industrial refrigeration viable and creating an engineering need to quantify cycle performance with COP.
1930s–1950s
Widespread Heat Pump Adoption
Heat pumps begin to appear in residential and commercial buildings. The COP metric becomes essential for comparing systems that can both heat and cool, guiding design and regulatory standards.

The central question this lesson addresses is straightforward yet powerful: given a device that consumes work to transfer heat between two thermal reservoirs, how do we measure the effectiveness of that transfer? Unlike thermal efficiency for heat engines — which is always less than one — the COP for refrigerators and heat pumps can be, and typically is, greater than unity, reflecting the fact that these devices leverage work to move energy rather than convert it.

Core Principles & Definitions

Before diving into calculations, it is essential to establish the foundational ideas that underpin the COP concept. Both refrigerators and heat pumps operate on reversed thermodynamic cycles — they consume net work input (Wnet) to transfer heat from a low-temperature reservoir (TL) to a high-temperature reservoir (TH). The distinction between the two devices lies solely in which thermal effect is desired.

1

Reversed Cycle

A thermodynamic cycle that receives net work input and transfers heat from a cold reservoir to a hot reservoir, the reverse of a heat engine. Both refrigerators and heat pumps employ reversed cycles.
2

Desired Output ≠ Work Output

COP is defined as the ratio of the desired thermal effect (QL for a refrigerator, QH for a heat pump) to the net work input Wnet. Because the desired effect is a heat transfer, COP can exceed 1.
3

Energy Conservation

By the First Law applied to a complete cycle, QH = QL + Wnet. This identity links the COP definitions for refrigerators and heat pumps.
4

Carnot Upper Bound

No real device can exceed the COP of a Carnot (fully reversible) cycle operating between the same two reservoirs. The Carnot COP depends only on TH and TL (in absolute units).
KEY TAKEAWAY
Think of a refrigerator or heat pump as a thermal elevator. Work is the electricity that powers the motor, but the payload is heat. Just as an elevator can lift a heavy load much higher than the energy stored in its motor's fuel, a heat pump can deliver far more thermal energy to a room than the electrical energy consumed — hence COP values greater than one. The COP tells you how many 'floors' of heat you move per unit of work.

Visual Explanation — Energy Flow Diagrams

Side-by-side energy flow diagrams for a refrigerator (left) and a heat pump (right). Both devices consume work W to transfer heat from TL to TH. The shaded box at the bottom of each diagram shows the COP definition: the refrigerator's desired output is QL (cooling), while the heat pump's desired output is QH (heating).

The diagram above captures the essential symmetry between a refrigerator and a heat pump. In both cases, the cycle device (typically a vapor-compression loop consisting of a compressor, condenser, expansion valve, and evaporator) absorbs heat QL from the cold reservoir and rejects heat QH to the hot reservoir, with energy conservation demanding QH = QL + Wnet. The only conceptual difference is which reservoir houses the space of interest. For the refrigerator it is the cold side (we want to keep it cold), and for the heat pump it is the hot side (we want to keep it warm). This distinction alone determines the numerator in the COP expression.

Mathematical Framework

We now formalize the COP definitions and derive the Carnot limits. Throughout, all heat quantities (QH, QL) and work (Wnet) are taken as positive magnitudes, following the convention most common in engineering thermodynamics texts.

COP — REFRIGERATOR
COP_R = Q_L / W_net
QL = heat removed from the cold reservoir (the refrigerated space); Wnet = net work input to the cycle. A higher COPR means more cooling per unit of work.
COP — HEAT PUMP
COP_HP = Q_H / W_net
QH = heat delivered to the hot reservoir (the heated space); Wnet = net work input. Because QH = QL + Wnet, COPHP is always greater than 1.
FUNDAMENTAL IDENTITY
COP_HP = COP_R + 1
Derived directly from the First Law: COPHP = QH / W = (QL + W) / W = QL / W + 1 = COPR + 1. This elegant result holds for every conceivable cycle, real or ideal.
CARNOT COP LIMITS
COP_R,Carnot = T_L / (T_H − T_L) ; COP_HP,Carnot = T_H / (T_H − T_L)
TH and TL are in absolute temperature units (K or R). These represent the maximum possible COP for any device operating between the two given reservoirs.

The Carnot COP expressions reveal a key physical insight: as the temperature difference (TH − TL) shrinks, both COPs increase and approach infinity in the limit TH → TL. Conversely, operating across a large temperature span demands more work per unit of heat transferred. Real cycles always have COP values below the Carnot values due to irreversibilities such as friction, heat transfer across finite temperature differences, and non-isentropic compression and expansion.

Carnot COP — Visualization & Interpretation

The Carnot COP for both refrigerators (cyan) and heat pumps (pink) plotted against the ratio TL/TH. As the ratio approaches 1 (small temperature difference), both COPs rise sharply. The yellow markers illustrate the identity COPHP = COPR + 1 at a representative point.

Several important observations emerge from this plot. First, the pink curve is always exactly one unit above the cyan curve, confirming the identity COPHP = COPR + 1 at every point. Second, both curves diverge as the temperature ratio TL/TH approaches unity, confirming the intuition that it is easier to move heat across a small temperature difference. At the extreme left where TL/TH → 0 (e.g., cooling toward absolute zero), COPR → 0 while COPHP → 1, reflecting the immense difficulty of extracting heat from very cold sources. These Carnot values serve as absolute upper bounds; any real device will fall below them.

Watch Your Units
Carnot COP formulas require temperatures in absolute units (Kelvin or Rankine). A common error is to plug in Celsius or Fahrenheit values. Always convert first: T(K) = T(°C) + 273.15 and T(R) = T(°F) + 459.67.

Worked Example — Refrigerator & Heat Pump COP

A household refrigerator maintains its interior at −5 °C while rejecting heat to a kitchen at 25 °C. The compressor consumes 0.6 kW of electrical power, and steady-state measurements indicate that 1.5 kW of heat is removed from the refrigerated space. Determine (a) the COP of the refrigerator, (b) the rate of heat rejection to the kitchen, (c) the COP if the same device operates as a heat pump delivering heat to the kitchen, and (d) the Carnot COP for both modes.

Refrigerator & Heat Pump COP Calculation
1
Step 1 — Identify Given ValuesTL = −5 °C = 268.15 K; TH = 25 °C = 298.15 K; Ẇnet = 0.6 kW; Q̇L = 1.5 kW.
2
Step 2 — Compute COP of the RefrigeratorCOPR = Q̇L / Ẇnet = 1.5 kW / 0.6 kW
COPR = 2.5
3
Step 3 — Find the Rate of Heat RejectionFirst Law for the cycle: Q̇H = Q̇L + Ẇnet = 1.5 + 0.6 = 2.1 kW
H = 2.1 kW
4
Step 4 — Compute COP as a Heat PumpCOPHP = Q̇H / Ẇnet = 2.1 / 0.6 = 3.5. Alternatively, COPHP = COPR + 1 = 2.5 + 1 = 3.5 ✓
COPHP = 3.5
5
Step 5 — Carnot COP ValuesCOPR,Carnot = TL / (TH − TL) = 268.15 / (298.15 − 268.15) = 268.15 / 30 ≈ 8.94. COPHP,Carnot = COPR,Carnot + 1 ≈ 9.94. The actual COPs (2.5 and 3.5) are well below the Carnot limits, as expected for a real device.
COPR,Carnot ≈ 8.94 ; COPHP,Carnot ≈ 9.94

Refrigerator vs. Heat Pump — Strengths & Limitations

Comparison of COP characteristics for refrigerators and heat pumps
AttributeRefrigerator (COP_R)Heat Pump (COP_HP)
Desired effectRemove heat from cold space (QL)Deliver heat to warm space (QH)
COP range0 < COPR < ∞ (typically 2–5 for domestic units)1 < COPHP < ∞ (typically 3–6 for residential systems)
Lower boundCOPR → 0 as TL → 0COPHP → 1 as TL → 0 (all work becomes QH)
Sensitivity to ΔTDecreases rapidly as TH − TL increasesSame sensitivity; COPHP drops as ΔT increases
Key irreversibilitiesCompressor friction, throttling losses, finite-ΔT heat exchangeSame irreversibilities; additionally, defrost cycles in air-source units reduce effective COP
KEY TAKEAWAY
A heat pump that uses the same cycle as a refrigerator will always have a COP that is exactly one unit higher. This is not an accident of engineering but a direct consequence of energy conservation: the heat delivered to the warm side includes both the heat extracted from the cold side and the work input. In the broader context of building energy systems, this means a heat pump can deliver, say, 3.5 kW of heating for every 1 kW of electricity, making it significantly more energy-effective than a simple electric resistance heater (which has, in effect, a COP of exactly 1).

Connection to Advanced Theory

The COP concept developed in this lesson assumes steady-state operation between two fixed-temperature reservoirs — the simplest idealization. Advanced coursework extends these ideas in several directions. Exergy analysis (also called availability analysis) moves beyond energy balances to account for the quality of energy, revealing where the greatest thermodynamic losses occur within the cycle. Second-law efficiencyII) compares a device's actual COP to the Carnot COP, providing a normalized metric of how close the device comes to theoretical perfection.

COP fundamentals vs. advanced thermodynamic analysis
ConceptThis Lesson (COP)Advanced Extension
Performance metricCOP = Desired Q / WnetηII = COPactual / COPCarnot
Reservoir modelTwo fixed-temperature reservoirsVariable-temperature sources/sinks, multi-stage cascades
Loss identificationCOP < COPCarnot signals irreversibility existsExergy destruction in each component pinpoints where losses occur
Cycle typesGeneric reversed cycle (Carnot as benchmark)Vapor-compression, absorption, gas-cycle (Brayton), thermoelectric

As you progress to courses in advanced thermodynamics or HVAC system design, you will encounter seasonal energy efficiency ratio (SEER) and heating seasonal performance factor (HSPF), which are essentially COP values averaged over an entire cooling or heating season and expressed in mixed imperial units (BTU/Wh). Understanding the fundamental COP definitions from this lesson provides the conceptual foundation for interpreting and comparing these industry-standard ratings.

Practice Problems

PROBLEM 1CONCEPTUAL
A friend claims that a heat pump with a COP of 0.8 is still useful because it 'moves heat uphill.' Explain why a COPHP less than 1 is thermodynamically impossible for a device operating as a heat pump. What does this imply about the relationship between QH and Wnet?
PROBLEM 2BASIC CALCULATION
A refrigerator removes 8 kJ of heat from a cold compartment per cycle, while the compressor requires 2.5 kJ of work per cycle. Determine (a) the COP of the refrigerator and (b) the heat rejected to the surroundings per cycle.
PROBLEM 3INTERMEDIATE
An air-source heat pump operates between outdoor air at −10 °C and a building interior at 22 °C. If the actual COPHP is 3.8, determine (a) the Carnot COPHP, (b) the Second-Law efficiency ηII = COPactual / COPCarnot, and (c) the power consumption if the heat pump must deliver 12 kW of heating.
PROBLEM 4APPLIED
A food-processing plant requires a refrigeration system that removes 200 kW of heat from a cold room at −25 °C. The condenser rejects heat to cooling water at 35 °C. Electricity costs $0.12 per kWh. If the system operates with COPR = 1.8 for 6000 hours per year, estimate (a) the annual electricity consumption in kWh, (b) the annual electricity cost, and (c) the savings if the COP could be improved to 2.5.
PROBLEM 5CRITICAL THINKING
Consider a Carnot refrigerator operating between TL and TH. Show that the Carnot COPR can be expressed as COPR,Carnot = (1/ηCarnot) − 1, where ηCarnot = 1 − TL/TH is the Carnot heat-engine efficiency operating between the same reservoirs. Discuss what this relationship reveals about the duality between heat engines and refrigerators.

Summary — COP for Refrigerators & Heat Pumps

The coefficient of performance (COP) measures how effectively a reversed thermodynamic cycle converts work input into a desired thermal effect. For a refrigerator, COPR = QL / Wnet, quantifying how much heat is removed from the cold space per unit of work. For a heat pump, COPHP = QH / Wnet, measuring how much heat is delivered to the warm space. The fundamental identity COPHP = COPR + 1 follows directly from the First Law of Thermodynamics.

The Carnot COP sets the theoretical upper limit: COPR,Carnot = TL / (TH − TL) and COPHP,Carnot = TH / (TH − TL), both requiring absolute temperature units. Real devices always fall below these limits due to irreversibilities such as friction, throttling, and finite-temperature-difference heat transfer. Comparing actual COP to Carnot COP via the Second-Law efficiency provides deeper insight into how much room remains for improvement.

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