Question

An immersion heater used to boil water for a single cup of tea plugs into a 120 V outlet and is rated at 450 W . Suppose your super-size, super-insulated tea mug contains 400 g of water at a temperature of 23 ∘C. You can ignore the energy needed to raise the temperature of the mug and the heater itself.
What is the resistance of the heater?

How long will this heater take to bring the water to a boil?

Answer 1

Answer:

Explanation:

Given

Voltage $V=120 V$

Power $P=450 W$

mass of water $m=400 gm$

initial temperature of water $T_i=23^{\circ}C$

Resistance R is given by

$P=\frac{V^2}{R}$

$R=\frac{V^2}{P}$

$R=\frac{120^2}{450}$

$R=32 \ Omega$

Heat required to raise water temperature to $100^{\circ}C$

$Q=mc\Delta T$

where $c=4.184 kJ/kg-K$ specific heat of water

$Q=0.4\times 4.184\times (100-23)=128.86 kJ$

time required is

$t=\frac{Q}{P}=\frac{128.86\times 1000}{450}$

$t=286.37 s\approx 4.77 min$