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 at 200 W Suppose your super-size, super-insulated tea mug contains 400 g of water at a temperature of 23 degree C. You can ignore the energy needed to raise the temperature of the mug and the heater itself. Part A What is the resistance of the heater? Express your answer to two significant figures and include the appropriate units. How long will this heater take to bring the water to a boil? Express your answer using two significant figures.

Explanation / Answer

Given

heater rated with power P = 200 W

when connected with V = 120 V

PArt A

resistance of the heater is

  
we know that at the heater the power dissipation will be in the form of heat energy absorbed by the water sourrounding the heater

so the formula for power dissipation is P = V^2/R ==> R = V^2/P = 120^2/200 ohm = 72 ohm

Part B

time taken for the heat to boil water from 23 0C is

first heat energy required is Q = mc*dT

               = 0.4*4128(100-23) J

               = 127142.4 J

now the power P= W/t ==> t = W/P

           t = 127142.4/200 s

           t = 635.712 s

           t = 10.5952 min

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