Heat flows from hot to cold on its own. A refrigerator moves it the other way, from the cold food to the warm kitchen. It is a heat engine run backward: work goes in, and heat is pumped uphill.
Why it works
Over each cycle $\Delta U = 0$, so energy in equals energy out: the heat $Q_C$ taken from the cold side plus the work $W$ equals the heat $Q_H$ given to the hot side. A refrigerator is rated by how much heat it removes per joule of work. A heat pump, the same machine pointed at a house, is rated by how much heat it delivers per joule, and that is always more than one joule, because it also brings in the heat it collected from outside.
Key idea
$$Q_H = Q_C + W, \qquad \text{COP}_{\text{fridge}} = \frac{Q_C}{W}, \qquad \text{COP}_{\text{heat pump}} = \frac{Q_H}{W}$$ The best possible (Carnot) values are $\dfrac{T_C}{T_H - T_C}$ and $\dfrac{T_H}{T_H - T_C}$, with kelvins.
Worked examples
Worked example: A kitchen refrigerator
A refrigerator removes $600\,\text{J}$ from the food using $200\,\text{J}$ of work. Find the heat released into the kitchen and the COP.
- $Q_H = Q_C + W = 600 + 200 = 800\,\text{J}$, released by the coils on the back.
- $\text{COP} = \dfrac{Q_C}{W} = \dfrac{600}{200} = 3.0$.
Answer: $800\,\text{J}$; COP $= 3.0$
Worked example: A heat pump
A heat pump with a heating COP of $3.5$ delivers $7.0\,\text{kW}$ of heat to a house. Find the electrical power and the rate it draws heat from outside.
- $W = \dfrac{Q_H}{\text{COP}} = \dfrac{7.0\,\text{kW}}{3.5} = 2.0\,\text{kW}$.
- From outside: $Q_C = Q_H - W = 7.0 - 2.0 = 5.0\,\text{kW}$.
Answer: $2.0\,\text{kW}$; $5.0\,\text{kW}$
Worked example: The best possible heat pump
What is the highest heating COP between $-5\,^\circ\text{C}$ outside and $20\,^\circ\text{C}$ inside?
- $T_H = 293\,\text{K}$, and the difference is $25\,\text{K}$.
- $\text{COP}_{\max} = \dfrac{293}{25} = 11.7$. Real heat pumps reach a fraction of this, and less in colder weather.
Answer: $11.7$
Common mistake
Thinking a heat pump that gives $3\,\text{J}$ of heat per joule of work breaks conservation of energy. It does not create heat; it moves $2\,\text{J}$ in from the cold outdoors and adds the $1\,\text{J}$ of work. Check: $Q_H = Q_C + W$ always balances.
Leaving a refrigerator door open warms the kitchen: it dumps $Q_H = Q_C + W$, which is more than the $Q_C$ it removes.