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:

32 Ω

286.50844 seconds

Explanation:

V = Voltage = 120 V

m = Mass of water = 400 g

c = Specific heat of water = 4186 J/kg°C

$\Delta T$ = Change in temperature = 100-23 (100°C boiling point)

P = Power = 450 W

R = Resistance

Power is given by

$P=\frac{V^2}{R}\\\Rightarrow R=\frac{V^2}{P}\\\Rightarrow R=\frac{120^2}{450}\\\Rightarrow R=32\ \Omega$

The resistance of the heater is 32 Ω

Heat generated

$Q=mc\Delta T\\\Rightarrow Q=0.4\times 4186\times (100-23)\\\Rightarrow Q=128928.8\ J$

Power is given by

$P=\frac{Q}{t}\\\Rightarrow t=\frac{Q}{P}\\\Rightarrow t=\frac{128928.8}{450}\\\Rightarrow t=286.50844\ s$

The time taken for the water to boil is 286.50844 seconds