Engines, refrigerators, and heat pumps

Why does a refrigerator warm the room?

Suppose a refrigerator removes 1200 J from its cold compartment in one cycle while receiving 400 J of work into its stated cycle. How much energy must its warm-side coils release into the room? Could opening the door cool the whole room? These questions begin with the energy crossing a chosen boundary, before any efficiency formula.

If the working fluid returns to its initial state each cycle, its stored energy has no net change. The warm side must therefore release 1600 J: the 1200 J removed from the compartment plus the 400 J input. The refrigerator moves energy from a colder region to a warmer one while adding its work input to the warm-side delivery.

Use assigned reservoir temperatures of 270 K and 300 K for this teaching case. A reservoir is an ideal source or sink that exchanges energy while retaining its temperature. A finite room or a cooling object is not exactly such a reservoir; its temperature can change. None of the assigned cases below is a measured device rating.

For a first reading, follow the energy directions, engine efficiency, the two coefficients of performance, reversible bounds, temperature lift and the four-component refrigerator. Sections marked “Further” offer a return path through explicit gas cycles, finite-conductance power and an entropy audit. Published preparation is “Physical systems, states, boundaries, and environments” and “Units, standards, and physical constants”. The needed energy, cycle and temperature terms are defined here; neighboring thermodynamics lessons are follow-on support, not assumed live prerequisites.

A cycle restores the working substance, not the surroundings

A working substance is the material whose state changes while an engine or refrigerator operates. A thermodynamic cycle returns that material to its initial state. Its internal energy U, a state quantity, therefore has ΔU = 0 over the full cycle, even though heat and work have crossed its boundary. Reservoirs and a load can have changed after it completes the cycle.

Use signed heat Q positive into the working substance and signed work W_by positive out. The closed-system first law is ΔU = Q − W_by, neglecting endpoint kinetic and gravitational changes. Over a cycle, net heat into that system equals net work out. A continuously flowing plant instead needs a control-volume balance, including the energy carried by entering and leaving mass.

To keep heat arrows readable, Q_H and Q_C below are nonnegative magnitudes of hot-side and cold-side exchanges, with their directions stated separately. They are not both signed heat inputs. An engine receives Q_H from the hot source and rejects Q_C to the cold sink. A refrigerator receives Q_C from the cold region and rejects Q_H to the warm region while receiving work W_in.

Energy per cycle is in joules. A rate, written with a dot or described as power, is in watts: 1 W = 1 J/s. Multiplying a per-cycle quantity by cycles per second gives its average rate only for the stated repeatable operation. Startup, stored-energy changes, defrost and changing loads may require a longer time-dependent balance.

An engine converts part of heat input into net work

Thermal efficiency is η = W_out/Q_H = 1 − Q_C/Q_H for a cyclic engine with the stated two heat exchanges and no other energy input. It compares useful net work to heat supplied, rather than gross expansion work before compression or pump costs are subtracted.

For a separate assigned engine, take Q_H = 2000 J and Q_C = 1200 J per cycle. Then W_out = 800 J and η = 0.400, or 40.0%. At 5.00 cycles/s, the heat-input rate is 10.0 kW, rejection is 6.00 kW and net output is 4.00 kW. The rates close the same balance as the per-cycle values.

A cold sink is part of the thermal-engine arrangement. A cyclic device cannot have as its only effect the extraction of heat from a single reservoir and its complete conversion into work. That second-law restriction is stronger than energy conservation. A battery or fuel cell has other forms of energy change; applying a two-reservoir heat-engine formula blindly to its entire input misidentifies the conversion.

Practical plant efficiency also depends on the declared boundary: shaft work, generator output and net exported electrical energy differ when generators, pumps, fans and other loads are included. A cogeneration plant supplies useful heat as well as work; adding both useful outputs can describe energy utilization, but does not turn all delivered heat into work or make its thermal efficiency the same number.

One reversed cycle, two useful outputs

A refrigerator is judged by the heat removed from the cold region. Its coefficient of performance is COP_R = Q_C/W_in. A heating heat pump is judged by heat delivered to the warm region: COP_HP = Q_H/W_in. For the same cycle, boundary and operating condition, Q_H = Q_C + W_in gives COP_HP = COP_R + 1.

The opening assigned cycle has COP_R = 1200/400 = 3.00 and COP_HP = 1600/400 = 4.00. A COP above one does not mean that more energy was created than supplied. The useful heating output includes energy taken from the colder source as well as the work input. COP is not a heat-to-work thermal efficiency.

At the assigned 2.00 cycles/s, this cycle removes 2.40 kW, receives 0.800 kW of work and rejects 3.20 kW. Cooling capacity means a rate of heat removal; it is not the same quantity as COP. Two devices can have equal COP but very different capacity, or equal capacity but different input power.

An open refrigerator door inside one otherwise closed room does not provide an external cold source. Heat it removes from room air is returned to that same room along with the electrical input, so its net effect adds that input to room energy under the stated boundary. In normal operation the insulated cold compartment is a distinct useful region; an air conditioner rejects heat outside the space it cools.

What two reservoir temperatures can tell us

A reversible process is an ideal limiting process that can be reversed while restoring both system and surroundings without a residual change. Quasistatic motion alone does not remove friction, a finite-temperature heat-transfer difference or throttling irreversibility. A reversible cycle between two fixed-temperature reservoirs provides a performance bound, not a blueprint for a real machine at a promised output rate.

For T_H > T_C > 0 on the absolute kelvin scale, the maximum engine efficiency is η_rev = 1 − T_C/T_H. Reversing the reversible exchange ratio Q_C/Q_H = T_C/T_H gives COP_R,rev = T_C/(T_H − T_C) and COP_HP,rev = T_H/(T_H − T_C). Actual devices with the same two-reservoir boundary cannot exceed these values.

For the assigned engine at 600 K and 300 K, η cannot exceed 0.500. Its assigned 0.400 lies below that bound. A proposed 2000 J input, 800 J rejection and 1200 J work would conserve energy but have efficiency 0.600, above the bound; that two-reservoir cyclic claim is rejected.

For the opening refrigerator at 270 K and 300 K, the reversible cooling and heating COPs are 9.00 and 10.0. Its assigned values 3.00 and 4.00 are below those bounds. Removing the same 1200 J would require at least 133.333 J of work in the reversible limit. Temperature ratios using 0 °C and 30 °C would be invalid; offsets do not cancel in ratios.

Why changing the working fluid cannot beat that bound

Imagine a proposed engine more efficient than a reversible engine between the same reservoirs. Run the reversible device backward, and scale it so that the two work transfers cancel. The combined arrangement would move heat from the cold reservoir to the hot one with no other net effect. The second law excludes that result. Interchanging the two reversible devices similarly shows why they must share the same limiting efficiency.

The argument concerns complete cycles and all their external effects. Cooling a finite hot body once, consuming fuel, draining a battery or changing another reservoir is an additional effect. Such cases are not counterexamples to the two-reservoir cyclic bound; they require the corresponding larger energy and state balance.

A different working fluid can greatly affect attainable temperatures, pressures, heat-transfer rates, safety, losses and practical performance. It does not change the reversible bound for the same external reservoir temperatures. Conversely, a gas cycle that exchanges heat over a range of temperatures does not automatically attain the Carnot efficiency evaluated from its hottest and coldest working-fluid states.

Temperature lift and a declared heating load

Temperature lift is the hot-to-cold temperature difference in the stated model. Raising T_H or lowering T_C increases that lift and lowers the reversible heat-pump COP. For a fixed useful heating rate, the corresponding reversible minimum work rate is W_min = Q_H(T_H − T_C)/T_H. Larger lift requires more minimum input for that same load.

Hold the useful hot side at 308.15 K and compare five colder-source temperatures. The table varies only the reversible bound, not an assumed actual COP curve. Reservoir temperatures are external references; a real refrigerant must usually be warmer than the heated medium in its condenser and colder than the source in its evaporator to transfer heat at a finite rate.

At the separate assigned operating point T_C = 273.15 K, a building receives 6.00 kW from a compressor cycle with heating COP 3.20. Compressor input is 1.875 kW and extracted source heat is 4.125 kW. Include an additional 0.200 kW auxiliary load stipulated to dissipate outside the heated envelope: wall input is 2.075 kW and whole-system useful heating COP is about 2.892.

Over one hour at those constant assigned conditions, wall energy is 2.075 kWh. The stipulated lossless resistance-heating comparator would use 6.00 kWh for the same useful heat. These are model calculations, not a seasonal rating or purchasing comparison. If auxiliary dissipation instead enters the heated space, or defrost changes delivered heat, the useful-output balance must change too.

How a vapor-compression loop moves heat

A common refrigerator or heat pump circulates a refrigerant through a compressor, condenser, expansion valve and evaporator. In the usual dry-compression arrangement, low-pressure vapor enters the compressor. Shaft work raises its pressure and generally its temperature enough that it can release energy to the warm side.

In the high-pressure condenser, the refrigerant rejects energy and commonly condenses to liquid; its temperature must support heat transfer toward the warmer useful load or room. The liquid then passes through a restriction to the low-pressure side. An approximately adiabatic valve with no shaft work and negligible kinetic and gravitational changes has nearly equal inlet and outlet enthalpy, not a useful work output.

At a suitable low pressure, part of the refrigerant can flash into vapor through the valve. The resulting cold stream absorbs energy in the evaporator and vaporizes before returning to the compressor. Pressure establishes the relevant phase-equilibrium temperature for a chosen fluid; compression does not simply “make cold”, and a valve is not an isentropic turbine.

The usual valve/evaporator cooling behavior depends on the refrigerant and its state. Isenthalpic does not mean universally cooling, and enthalpy values alone do not establish pressure, temperature or vapor quality. Superheating, subcooling, pressure drops and real compressor performance change the detailed cycle. A reversible thermodynamic cycle is not obtained simply by reversing the motor of this irreversible loop.

Use enthalpy for the flowing components

For steady flow, mass carries internal energy and the pressure–volume contribution needed to enter or leave a component. Their sum per mass is specific enthalpy h = u + pv, with u in J/kg, pressure p in Pa and specific volume v in m³/kg. Under the stated negligible kinetic/gravitational changes, the component energy balance relates heat and shaft work to enthalpy differences.

Use a separate synthetic ledger: h₁ = 400, h₂ = 450, h₃ = 250 and h₄ = 250 kJ/kg at suction, compressor outlet, condenser outlet and valve outlet. The compressor requires h₂−h₁ = 50 kJ/kg; the condenser releases h₂−h₃ = 200 kJ/kg; the evaporator absorbs h₁−h₄ = 150 kJ/kg. Their equality 200 = 150 + 50 is the same cycle balance as before.

At an assigned mass flow of 0.0200 kg/s, these become 1.00 kW input, 4.00 kW warm-side delivery and 3.00 kW cooling. The cooling and heating COPs are 3.00 and 4.00 for this ledger. Multiply kJ/kg by kg/s and by 1000 to obtain W; overlooking that conversion gives a thousandfold error.

These supplied h values are not a property table for a named fluid and do not validate a physical pressure/phase design. Actual analysis also checks refrigerant properties, pressure levels, dry compressor inlet, material and operating limits, energy leaks and component efficiencies. A p–v loop area for a closed piston gas is not automatically the shaft work of a continuously flowing compressor circuit.

What a real performance figure includes

Real heat exchangers need driving temperature differences; friction and pressure drops dissipate work, and nonideal compression raises required input. Fans, pumps, controls, crankcase heating, cycling, startup and frosting or defrost can also matter at the declared system boundary. A single compressor-cycle COP does not by itself describe all energy purchased or all heat delivered over a season.

The heating or cooling capacity may change with source temperature and operating limits. A large COP at very small ideal temperature lift does not imply unlimited cooling power or finite heat transfer across zero temperature difference. The useful rate depends on heat-exchanger size, driving differences, flow and the specific machine.

A seasonal ratio is generally total useful energy delivered divided by total relevant input energy over its specified conditions. It is not the unweighted arithmetic mean of point COPs. Test procedures may use different units and include different loads, so convert units and inspect the declared conditions before comparing figures.

For air-source operation, frosting can obstruct the outdoor heat exchanger and defrost can temporarily change power and useful output. Ground- and water-source arrangements have different source conditions and auxiliary demands. These mechanisms explain why a theoretical bound and an assigned single operating point are useful checks but not substitutes for a full system measurement.

Different cycle families answer different physical questions

A Carnot cycle specifies reversible isothermal heat exchange connected by reversible adiabatic changes. It supplies the two-reservoir benchmark. An ideal Otto cycle instead represents compression, rapid heat addition near fixed volume, expansion and fixed-volume heat rejection of a closed gas surrogate. A real spark-ignition engine also exchanges mass and releases chemical energy, so that surrogate needs explicit limits.

The ideal Brayton gas-turbine model uses reversible adiabatic compression and expansion connected by approximately constant-pressure heat addition and rejection. Compressor input must be subtracted from turbine output. A real open gas turbine has intake and exhaust; its closed working-fluid cycle is a modeling construction, with the environment providing the return process.

A Rankine vapor-power model uses liquid pumping, high-pressure heating/vaporization, turbine expansion and low-pressure condensation. Pumping a dense liquid and expanding vapor differ from compressing a gas throughout. Phase properties, turbine wetness, pressure losses and heat-source temperatures shape its useful operating regime; it is not a Carnot steam engine simply because it uses boiling and condensation.

Refrigeration can also use an expansion device that recovers work or a reversed gas cycle when the application warrants it. Heat-driven absorption systems use an additional thermal input and a solution/regeneration mechanism; their thermal COP denominator differs from the electrical-work denominator used here. Include every generator, absorber, pump and heat exchange before applying a balance or comparing ratios. Detailed engineering and phase/potential models belong to their respective follow-on subjects.

Further: follow all four states of a Carnot gas cycle

For an explicit original realization, choose n = 1.00 mol of ideal gas with pV = nRT and R = 8.31446261815324 J/(mol K). Here C_p and C_V are total heat capacities at constant pressure and volume. Assign their constant ratio γ = C_p/C_V = 1.40; the ideal-gas relation C_p−C_V = nR gives C_V = nR/(γ−1), with ΔU = C_VΔT for this caloric model. Work is frictionless reversible pressure–volume work W_by = ∫p dV. This interpretation requires those conditions, rather than a curve drawn through arbitrary irreversible endpoints.

At 600 K, A→B expands isothermally from 0.0100 to 0.0200 m³ and absorbs Q_H = nRT_H ln 2 ≈ 3457.888 J. Reversible adiabatic B→C lowers the gas to 300 K; with constant γ, TV^(γ−1) is constant. It reaches V_C ≈ 0.113137 m³. Isothermal C→D at 300 K releases about 1728.944 J while compressing to V_D ≈ 0.056569 m³.

Adiabatic D→A completes the state cycle at 600 K and 0.0100 m³. The two adiabatic work transfers are equal and opposite for this assigned ideal gas because their temperature changes reverse. Net work is about 1728.944 J, and η = 0.500. The four branch changes individually obey ΔU = Q − W_by and sum to zero in U.

The clockwise p–V loop has net work equal to its signed enclosed area under these quasistatic pressure–volume conditions. Reversing all four reversible branches gives net input work and reversed heat exchanges. Isotherms need reversible thermal contact with their reservoirs, while the adiabats need thermal isolation. The detailed entropy lesson explains why reversibility is the decisive restriction.

Further: an Otto model and its conditional compression-ratio trend

For a separate one-mole constant-γ ideal-gas surrogate, start at T₁ = 300 K and V₁ = 0.0100 m³. Let compression ratio r = V₁/V₂ = 8.00 and peak T₃ = 1800 K, with γ = 1.40. The reversible adiabats give T₂ = T₁r^(γ−1) ≈ 689.219 K and T₄ = T₃/r^(γ−1) ≈ 783.496 K.

Fixed-volume heat addition is Q_H = C_V(T₃−T₂) ≈ 23088.867 J and rejection magnitude is Q_C = C_V(T₄−T₁) ≈ 10050.013 J. Net work is about 13038.854 J. Substituting both adiabatic relations gives η_Otto = 1−r^(1−γ) ≈ 0.564725.

This efficiency depends on r and the assigned constant γ within this model. It does not say that a real engine can increase compression ratio indefinitely: ignition/knock, material limits, changing heat capacities, combustion timing, heat loss, friction and gas exchange matter. Nor is it a comparison with the 600/300 K Carnot example; this surrogate has different temperatures and heat-exchange conditions.

During the Otto constant-volume heat exchanges, working-fluid temperature changes. A reversible realization would require matched changing-temperature heat sources/sinks rather than one hot and one cold isotherm. The paired cycle figures therefore retain separate pressure and volume scales and do not compare loop areas as if they belonged to the same device or energy input.

Further: finite heat transfer can separate maximum power from maximum efficiency

Consider a bounded original model: a reversible internal engine works between θ_H and θ_C, while external reservoirs are at 600 K and 300 K. Heat crosses two finite linear conductances K_H = K_C = 10.0 W/K. Thus Q̇_H = K_H(T_H−θ_H) and Q̇_C = K_C(θ_C−T_C). The internal reversible ratio is Q̇_C/Q̇_H = x = θ_C/θ_H; this does not make the external heat transfers reversible.

Solving the two balances gives θ_H = (K_HT_H + K_CT_C/x)/(K_H+K_C) and θ_C = xθ_H. With K_eff = K_HK_C/(K_H+K_C), net power is P = K_eff(1−x)(T_H−T_C/x), valid for T_C/T_H < x < 1. Efficiency is 1−x. These equations define a particular steady or cycle-averaged model, not every finite-time engine.

Differentiation gives dP/dx = K_eff(−T_H + T_C/x²). Its interior maximum is x = √(T_C/T_H). Here θ_H ≈ 512.132 K, θ_C ≈ 362.132 K, maximum power ≈ 257.359 W and efficiency there ≈ 0.292893. The same model approaches the Carnot efficiency 0.500 as the heat-transfer driving differences and power approach zero.

The expression 1−√(T_C/T_H) is therefore an efficiency at maximum power for these stated conductance and internal-reversibility assumptions. It is not a new universal upper bound on all real engine efficiencies. Different losses, transfer laws, constraints or optimization objectives can give different results. This example explains why “best efficiency” and “largest useful rate” are different design questions.

Further: a compact second-law audit of a cycle

Entropy S is a state quantity used to track a second-law balance. A cyclic working substance has ΔS = 0. For the device plus two constant-temperature reservoirs and no additional uncounted effects, reservoir entropy changes give total entropy generation S_gen = Q_C/T_C − Q_H/T_H for an engine, or S_gen = Q_H/T_H − Q_C/T_C for a refrigerator. It must be nonnegative, with zero in the globally reversible limit.

For the assigned engine, S_gen = 1200/300 − 2000/600 ≈ 0.666667 J/K per cycle. The rejected candidate with Q_C = 800 J instead gives about −0.666667 J/K. The negative value shows that its energy-conserving ledger is incompatible with the stated second-law boundary.

For the opening refrigerator, S_gen = 1600/300 − 1200/270 ≈ 0.888889 J/K per cycle. Setting the generation to zero recovers the reversible heat ratio and the COP bounds. A nonnegative value is a necessary audit; it does not by itself prove that suitable materials, pressures, phase states, controls or heat-exchanger rates can realize a proposed machine.

Finite sources, extra reservoirs, material flows, heat leaks and changing stored energy require their corresponding entropy and energy terms. Do not substitute a working-fluid temperature for the reservoir temperature without identifying the thermal boundary. “Entropy and the second law” develops this framework; this return section applies only the bounded cyclic audit needed to evaluate the performance claims here.

Check a performance claim from its physical boundary

First identify the useful output: net work, cooling heat removed or heating heat delivered. Draw the system boundary and all heat, work and mass exchanges. Specify whether the claim concerns one complete cycle, steady operation, startup or a season, and whether temperatures describe reservoirs, refrigerant states or a changing finite load.

Close the energy balance with consistent signs, units and counting basis. Form η or the appropriate COP using that same boundary, then compare only with a limit whose assumptions match it. Check kelvin temperatures, auxiliary loads, heat-source changes and whether the stated power or capacity is also physically supported.

A useful model can be simple: the opening refrigerator has one cold-side input, one work input and one larger warm-side output. More detailed component and gas-cycle models explain where those exchanges occur and why losses matter. Their strength comes from retaining conditions, rather than treating a convenient efficiency formula as the machine itself.

Worked examples

Try it yourself

1. Does ΔU=0 over a cycle mean there was no heat or work?

   Show the answer and reasoning

   No.

   It means the final state energy equals the initial state energy. Nonzero heat and work transfers can balance.

2. For an engine, are both positive magnitudes Q_H and Q_C heat into the working substance?

   Show the answer and reasoning

   No.

   Q_H enters, while Q_C leaves. The signed cycle heat is Q_H−Q_C.

3. Does COP_HP=4 violate energy conservation?

   Show the answer and reasoning

   No.

   Delivered heating includes cold-source energy and work input. COP is not a heat-to-work efficiency.

4. Can one always add one to a catalog cooling COP to predict a separate heating test COP?

   Show the answer and reasoning

   No.

   COP_HP=COP_R+1 requires the same cycle, operating point and boundaries. Different tests and auxiliaries change the comparison.

5. Can an open electric refrigerator cool the whole otherwise closed room containing it?

   Show the answer and reasoning

   No under the stated boundary.

   Its internal heat redistribution cancels in the room balance; its work input adds room energy.

6. May 0 °C and 30 °C be inserted directly into the Carnot COP ratio?

   Show the answer and reasoning

   No.

   Absolute-temperature ratios require kelvin. Celsius offsets cancel in differences but not ratios.

7. The assigned refrigerant loop removes 3000 W from the cold region while receiving 1000 W of shaft work. Find its cooling COP and warm-side heat rate. Which value describes its cooling capacity?

   Show the answer and reasoning

   COP_R = 3.00, warm-side delivery = 4000 W, and cooling capacity = 3000 W.

   Use the same steady loop boundary: COP_R = 3000/1000 = 3.00 and Q̇_H = Q̇_C + Ẇ_in = 3000 + 1000 = 4000 W. The dimensionless COP is a ratio; the 3000 W cooling capacity is a heat-removal rate.

8. Does reversing a compressor motor make a valve-based refrigerator reversible?

   Show the answer and reasoning

   No.

   The actual loop has component roles and irreversibilities. A reversible reverse cycle requires all changes and external effects to reverse without residual change.

9. Does an approximately adiabatic throttle with h₃=h₄ deliver turbine shaft work?

   Show the answer and reasoning

   No under the stated conditions.

   The valve has no shaft-work term and is generally irreversible. Equal enthalpy does not mean equal entropy or universal cooling.

10. Do the four supplied synthetic h values prove a dry vapor compressor inlet?

    Show the answer and reasoning

    No.

    Pressure, temperature, composition and phase/property information are also required. The ledger establishes only its stated energy accounting.

11. Does diverging reversible COP as lift tends to zero imply unlimited cooling capacity?

    Show the answer and reasoning

    No.

    That ideal ratio does not guarantee a finite driving temperature difference, heat-transfer rate or machine capacity.

12. Would the building example’s useful COP stay exactly 6000/2075 if the 200 W auxiliary heat instead entered the room and counted as useful delivery?

    Show the answer and reasoning

    No.

    The useful-output boundary would change; its 6000 W stipulated condenser output must be distinguished from additional indoor dissipation.

13. Does the assigned Otto efficiency exceed the assigned Carnot example at the same reservoir conditions?

    Show the answer and reasoning

    That comparison is not at the same conditions.

    The Otto surrogate has a different temperature range and variable-temperature heat exchange. The examples retain separate assumptions.

14. Is 1−√(T_C/T_H) a universal efficiency limit for all engines?

    Show the answer and reasoning

    No.

    It is the efficiency at maximum power for the stated endoreversible linear-conductance model; other constraints and loss laws can differ.

15. Does S_gen≥0 alone prove a proposed refrigerator can be built with the specified phase states and capacity?

    Show the answer and reasoning

    No.

    It is a necessary second-law check. Property, material, component, rate and control feasibility still require evidence.

16. Is a seasonal COP generally an unweighted arithmetic mean of point COPs?

    Show the answer and reasoning

    No.

    It is formed from total useful energy and total relevant input energy under the declared conditions; operating durations and loads matter.

What to carry forward

• A cyclic working substance returns to its state while heat, work and reservoir changes remain.
• An engine delivers Q_H−Q_C as net work; a refrigerator/heat pump receives that work difference to move heat uphill.
• COP_R and COP_HP count different useful outputs; their difference is one only for the same cycle and boundary.
• Carnot limits use absolute reservoir temperatures and globally reversible operation; they are not measured performance or finite-capacity predictions.
• Vapor-compression and ideal gas cycles explain distinct mechanisms, with stated enthalpy, phase, pressure–volume and caloric assumptions.
• Auxiliaries, temperature lift, losses, heat-transfer rates and complete energy/second-law boundaries decide what a performance figure means.