Problem — Heat pump between 295 and 246 K

Problem

A heat pump operates between an indoor temperature of $295\;\text{K}$ and an outdoor one of $246\;\text{K}$ (Carnot model). Calculate the coefficient of performance (COP). How much electrical energy is needed to deliver $5{,}0\;\text{kJ}$ of heat indoors?

Solution

COP of the heat pump (Carnot model). With $\text{COP}_\text{pdc} = Q_H/L$ and, in the reversible limit, $|Q_C|/Q_H = T_C/T_H$:

$\text{COP}_\text{pdc} = \frac{T_H}{T_H - T_C} = \frac{295}{295 - 246} = \frac{295}{49} = \ev{6{,}0}$

Electrical energy needed. From $\text{COP}_\text{pdc} = Q_H/L$ we get $L = Q_H/\text{COP}_\text{pdc}$, with $Q_H = 5{,}0\;\text{kJ}$:

$L = \frac{5{,}0\;\text{kJ}}{6{,}0} = \ev{0{,}83\;\text{kJ}}$

Only about $0{,}83\;\text{kJ}$ of electrical energy is needed to supply $5{,}0\;\text{kJ}$ of heat: the remaining $\approx 4{,}2\;\text{kJ}$ are taken for free from the outdoor air. The pump does not create energy, it moves it.

Links

Topics: Macchine termiche Concepts: Frigorifero e pompa di calore · Ciclo di Carnot Skills: Impostazione simbolica Objects: Frigorifero