The second law: heat engines, refrigerators and Carnot

The first law says that energy is conserved, so an engine cannot produce more work than the heat it takes in. It does not say that an engine cannot turn all of that heat into work. Yet every engine ever built has rejected some heat to its surroundings: a car's radiator and exhaust, a power station's cooling towers, the condenser of Watt's steam engine. The second law of thermodynamics makes this observation into a principle. This lesson states the second law in the two classical forms due to Kelvin and Clausius, shows that they are equivalent, proves Carnot's theorem that no engine can beat a reversible one, derives the Carnot efficiency, and applies it to engines, refrigerators and the heat pumps that are increasingly used to heat homes in cold climates.

Heat engines

A heat engine is a device that works in a cycle, taking in heat \(Q_h\) from a hot reservoir at temperature \(T_h\), doing work \(W\), and rejecting heat \(Q_c\) to a cold reservoir at \(T_c\). A reservoir is a body so large that its temperature does not change when heat flows in or out: a furnace, the atmosphere, a lake. Because the engine returns to its initial state after each cycle, its internal energy is unchanged, and the first law gives \(W = Q_h - Q_c\). The efficiency is the fraction of the heat input that becomes work,

Here \(Q_h\) and \(Q_c\) denote magnitudes. A refrigerator or heat pump runs the cycle backwards: work \(W\) is used to extract heat \(Q_c\) from the cold reservoir and deliver \(Q_h = Q_c + W\) to the hot one. A refrigerator is rated by its coefficient of performance \(\text{COP}_R = Q_c/W\); a heat pump, whose purpose is to warm the hot side, by \(\text{COP}_{HP} = Q_h/W = 1 + \text{COP}_R\).

Two statements of the second law

Kelvin, in 1851, and Max Planck later, stated the second law as an impossibility for engines:

Kelvin–Planck statement: no device operating in a cycle can take in heat from a single reservoir and convert it entirely into work.

Clausius, in 1854, stated it as an impossibility for refrigerators:

Clausius statement: no device operating in a cycle can transfer heat from a colder body to a hotter body with no other effect.

The two statements are equivalent: if either were false, the other would be too. Suppose a perfect engine existed that turned heat \(Q\) from a hot reservoir entirely into work \(W = Q\). Use that work to drive an ordinary refrigerator that takes \(Q_c\) from a cold reservoir and delivers \(Q_c + W\) to the hot one. The combination takes \(Q_c\) from the cold reservoir and delivers a net \(Q_c\) to the hot one, with no other effect, violating Clausius. Conversely, suppose a perfect refrigerator existed that moved heat \(Q_c\) from cold to hot with no work. Combine it with an ordinary engine that takes \(Q_h\) from the hot reservoir, rejects exactly \(Q_c\) to the cold one and does work \(Q_h - Q_c\). The cold reservoir is unchanged, and the combination turns the net heat \(Q_h - Q_c\) from the hot reservoir entirely into work, violating Kelvin–Planck. The second law is therefore one law, and the two forms are two of its faces.

The Carnot cycle

Carnot imagined an engine that uses only reversible processes. Its cycle has four steps, which for an ideal gas are: an isothermal expansion at \(T_h\), during which the gas absorbs \(Q_h\); an adiabatic expansion that cools the gas to \(T_c\); an isothermal compression at \(T_c\), rejecting \(Q_c\); and an adiabatic compression that returns it to \(T_h\). Heat flows only across vanishingly small temperature differences, and every step can be run backwards, so the same cycle run in reverse is a reversible refrigerator.

For an ideal gas the efficiency can be computed directly. On the isotherms, \(Q_h = nRT_h \ln(V_2/V_1)\) and \(Q_c = nRT_c \ln(V_3/V_4)\). On the adiabats, \(T_hV_2^{\gamma-1} = T_cV_3^{\gamma-1}\) and \(T_hV_1^{\gamma-1} = T_cV_4^{\gamma-1}\). Dividing these gives \(V_2/V_1 = V_3/V_4\), so the logarithms are equal and

Carnot's theorem

Carnot's theorem states that no engine operating between two reservoirs can be more efficient than a reversible engine operating between the same reservoirs, and that all reversible engines between them have the same efficiency. The proof is another combination argument. Suppose some engine X had efficiency greater than a reversible engine C. Run C backwards as a refrigerator, driven by the work of X, and adjust the sizes so that X takes in the same heat \(Q_h\) from the hot reservoir that C returns to it. Since X is more efficient, it produces more work than C needs; the hot reservoir is unchanged, and the surplus work has been extracted from the cold reservoir alone, violating Kelvin–Planck. If two reversible engines had different efficiencies, the same argument run with either one backwards would give a contradiction, so they must be equal.

The theorem has a striking consequence. Because every reversible engine between two reservoirs has the same efficiency, whatever its working substance, the ratio \(Q_c/Q_h\) for a reversible engine depends only on the two reservoirs and can be used to define a temperature scale independent of any substance. Kelvin proposed exactly this: define the ratio of two thermodynamic temperatures by \(T_c/T_h = Q_c/Q_h\) for a reversible engine between them. Our ideal gas calculation shows that this thermodynamic temperature coincides with the ideal-gas temperature of lesson 2, so the kelvin scale is both practical and fundamental. Absolute zero is the temperature of a cold reservoir to which a reversible engine would reject no heat at all.

Real engines

Real engines are less efficient than Carnot's, for two kinds of reasons. First, their cycles are not Carnot cycles: heat is absorbed over a range of temperatures, not all at the hottest. Second, they are irreversible: friction, turbulence and heat flow across finite temperature differences all waste some of the potential work. Engineers model them with idealized air-standard cycles. In the Otto cycle of a spark-ignition engine, heat is added and removed at constant volume, and the efficiency depends only on the compression ratio \(r\),

which is about 60% for \(r = 10\) and \(\gamma = 1.4\); real gasoline engines achieve 25% to 35% at their best operating point. The diesel cycle adds heat at constant pressure and uses higher compression ratios, typically 15 to 20; large marine diesels reach about 50%. A modern combined-cycle gas power plant, which uses the hot exhaust of a gas turbine to raise steam for a second turbine, reaches about 60%, close to the Carnot limit for its temperatures times a realistic factor. The Stirling cycle, invented by Robert Stirling in 1816, exchanges heat at constant temperature and stores the heat released in one constant-volume step in a regenerator, returning it in the other; with a perfect regenerator its efficiency equals Carnot's.

Refrigerators and heat pumps

For a reversible refrigerator, \(Q_c/Q_h = T_c/T_h\) as before, so

Both grow without limit as the temperature difference shrinks. This is the great advantage of heat pumps: moving heat across a small temperature difference costs much less work than creating it from electricity in a resistance heater, which has a COP of exactly 1. A heat pump warming a house from mild outdoor air can deliver three or four units of heat per unit of electricity. As the outdoor temperature falls, the COP falls too, both because the temperature lift grows and because the outdoor coil must be colder than the air in order to absorb heat from it. Cold-climate heat pumps, designed with variable-speed compressors and refrigerants suited to low temperatures, are now commonly rated to keep working down to about \(-25\) to \(-30\) °C, and they are being installed widely across the Prairies and the northern United States.

import math
gamma = 1.4

def carnot(Th, Tc): return 1 - Tc / Th
def otto(r): return 1 - r**(1 - gamma)
def diesel(r, rc): return 1 - r**(1 - gamma) * (rc**gamma - 1) / (gamma * (rc - 1))

print("engine efficiencies")
print(f"  Carnot between 900 K and 300 K:        {carnot(900, 300):.3f}")
for r in (8, 10, 12):
    print(f"  ideal Otto, compression ratio {r:2d}:      {otto(r):.3f}")
print(f"  ideal Diesel, r = 18, cutoff ratio 2:  {diesel(18, 2):.3f}")
print(f"  ideal Stirling with perfect regenerator = Carnot between its two temperatures")

# Heat pump heating a house to 21 C as the outdoor temperature falls
T_in = 294.15
print("\noutdoor (C)  Carnot COP  realistic COP (45 % of Carnot, 10 K approach)")
for t_out in (10, 0, -10, -20, -25, -30, -35):
    Tc = t_out + 273.15
    cop_carnot = T_in / (T_in - Tc)
    Th_coil, Tc_coil = T_in + 10, Tc - 10              # coil temperatures differ from the air
    cop_real = 0.45 * Th_coil / (Th_coil - Tc_coil)
    print(f"   {t_out:5d}       {cop_carnot:6.2f}        {cop_real:5.2f}")

# A refrigerator: kitchen at 22 C, freezer at -18 C, 1 kW of electrical input
Th, Tc = 295.15, 255.15
cop_ref = Tc / (Th - Tc)
print(f"\nfreezer: Carnot COP = {cop_ref:.2f}; with 1.0 kW input it could remove at most "
      f"{cop_ref:.1f} kW and dump {cop_ref + 1:.1f} kW into the kitchen")

engine efficiencies
  Carnot between 900 K and 300 K:        0.667
  ideal Otto, compression ratio  8:      0.565
  ideal Otto, compression ratio 10:      0.602
  ideal Otto, compression ratio 12:      0.630
  ideal Diesel, r = 18, cutoff ratio 2:  0.632
  ideal Stirling with perfect regenerator = Carnot between its two temperatures

outdoor (C)  Carnot COP  realistic COP (45 % of Carnot, 10 K approach)
      10        26.74         4.42
       0        14.01         3.34
     -10         9.49         2.68
     -20         7.17         2.24
     -25         6.39         2.07
     -30         5.77         1.93
     -35         5.25         1.80

freezer: Carnot COP = 6.38; with 1.0 kW input it could remove at most 6.4 kW and dump 7.4 kW into the kitchen

Even a Carnot heat pump loses ground in the cold, falling from a COP of about 27 at 10 °C to 5.8 at \(-30\) °C. The realistic model, which takes 45% of the Carnot value between coil temperatures 10 K beyond the indoor and outdoor air, gives COPs from 4.4 down to 1.9, close to what manufacturers report for good cold-climate units. At −30 °C (−22 °F), a very cold Minneapolis night, the heat pump still delivers almost twice as much heat per kilowatt-hour as a baseboard heater, though many installations keep a backup heater for the coldest hours, when the house also needs the most heat. The freezer calculation shows the other side of the coin: a refrigerator with 1 kW of input can remove at most about 6.4 kW from the freezer, and every joule of work it uses ends up as heat in the kitchen.