An immersion heater used to boil water for a single cup of tea plugs into a 120

Question

An immersion heater used to boil water for a single cup of tea plugs into a 120 V outlet and is rated at400 W . Suppose your super-size, super-insulated tea mug contains 400 g of water at a temperature of 18 C. You can ignore the energy needed to raise the temperature of the mug and the heater itself.

A. What is the resistance of the heater?Express your answer to two significant figures and include the appropriate units.

B. How long will this heater take to bring the water to a boil?Express your answer using two significant figures.

Explanation / Answer

(a) Power rating = Voltage^2/Resistance

P = V^2/R

400 = 120^2/R

R = 36 ohms

(b) Heat required to boil water = Power x time

mass x specific heat capacity of water x temperature difference = Power x time

mc( Tf - Ti) = Pt

0.4 x 4200 ( 100 - 18) = 400 x t

t = 344.4 s = 3.4 x 10^2 s

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