Efficiency of a kettle

Problem

A $500\ \text{W}$ electric heater brings $1{,}5\ \text{L}$ of water from $20\ ^\circ\text{C}$ to $80\ ^\circ\text{C}$ in $14\ \text{minutes}$. Calculate the efficiency and explain where the rest of the energy goes.

Solution

Useful energy. This is the heat actually absorbed by the water ($m = 1{,}5\ \text{kg}$, $\Delta T = 60\ \text{K}$):

$Q = m\,c\,\Delta T = 1{,}5\cdot 4186\cdot 60 = 376740\ \text{J}$

Energy supplied. This is the electrical energy delivered in $t = 14\cdot 60 = 840\ \text{s}$:

$E = P\,t = 500\cdot 840 = 420000\ \text{J}$

Efficiency.

$\eta = \frac{Q}{E} = \frac{376740}{420000} = 0{,}897$

Where the rest goes. About $10\%$ of the energy is dissipated: it heats the container and the heating element, transfers to the surrounding air by conduction and convection, and some is lost through evaporation of the water.

$\ev{\eta\approx 89{,}7\%}$

Links

Topics: Thermodynamics Concepts: Specific heat and heat capacity · Efficiency