Skip to content

Heat pumps

Heat and matter · level 8, lesson 3 of 6

A question to start

On Earth, a heat pump can give a house three or four joules of heat for every joule of electricity it uses. Could one heat a Mars home as cheaply, pulling heat from air at −59 °C?

The idea

A heat pump is a fridge turned round: it uses work \(W\) to take heat \(Q_c\) from a cold place and deliver heat \(Q_h = Q_c + W\) to a warm one. Its coefficient of performance for heating is

\[\text{COP} = \frac{Q_h}{W}.\]

The second law sets the best possible value, for a perfect pump between \(T_h\) and \(T_c\) in kelvin:

\[\text{COP}_\text{max} = \frac{T_h}{T_h - T_c}.\]

The bigger the temperature gap, the lower the COP. Real heat pumps reach only part of this, often a quarter to a half.

warm roomcold outsidepump heat outheat inwork W
Best possible COP, room at 20 °C (293 K) Mars Earth
Outside −59 °C (214 K), the average suppose 0 °C (273 K) in winter
\(T_h - T_c\) 79 K 20 K
\(\text{COP}_\text{max}\) 3.7 14.7

Worked examples

1. On Mars. \(293/(293 - 214) = 293/79 = 3.7\). Even a perfect pump gives under 4 J of heat per joule of work.

2. A better cold side. Suppose the pump draws heat from a reactor's warm radiator loop at 330 K instead, to heat water to 340 K: \(\text{COP}_\text{max} = 340/10 = 34\). Waste heat makes a far better source than Mars air.

3. The heater it replaces. An electric heater has a COP of exactly 1: each joule of electricity gives one joule of heat. Suppose a real Mars heat pump reaches half the best possible, a COP of about 1.8. It would need \(1/1.8 = 56\%\) of the electricity a heater needs.

Try it

  1. Find \(\text{COP}_\text{max}\) for heating a room at 293 K from air at 253 K.
  2. A heat pump with COP 3 uses 1 kW of electricity. How much heat does it give? How much does it take from outside?
  3. Why does a heat pump struggle more on Mars than on Earth?
  4. Why can a heat pump's COP be more than 1 without breaking the first law?
Answers
  1. \(293/40 = 7.3\).
  2. 3 kW of heat, of which 2 kW is taken from outside.
  3. The temperature gap is far bigger, which lowers the best possible COP.
  4. It does not make the heat; it moves heat from outside and adds the work. Energy out = heat taken + work.

Common mistake

Using °C. \(20/(20 - (-59)) = 0.25\) is nonsense. In kelvin: \(293/79 = 3.7\).

Where it leads