(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The heating element in a kettle has a resistance of 10 ohms when it is plugged into a 110V outlet. When heating water, only 75% of the energy produced is actually absorbed by the water. How much heat energy does the water absorb if it takes 4 minutes to boil the water?

2. Relevant equations

V = IR

P = VI

Percent Efficiency = Useful Energy Output / Total Electrical Output

3. The attempt at a solution

1. Finding Energy Produced by Kettle

I = V / R

I = 11a

P = VI

=110 x 11

=1210 w

290 400 j (1210 x 60 x 4) of energy produced in 4 minutes

290 400j x 0.75

=217 800j / 3600

=60.5 watt hours is absorb by the water.

However, my teacher wrote this on the board:

E = V x I x t

=110 x 11 x 240 (4 minutes in seconds)

=80 2/3 watt hours (output)

% Efficiency = Useful Energy Output / Total Electrical Input

75% = 80 2/3 / Total Electrical Energy Input

E = 80 2/3 divided by 0.75

E absorbed by water = 108 watt hours

His answer confuses me... how could the energy absorbed by the water be more then the energy output?

Please help, have a "quest" on this tomorrow.

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# Percent Efficieny Problem

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