# Percent Efficieny Problem

1. Mar 8, 2009

### lnvoker

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?

2. Mar 8, 2009

### Delphi51

Teachers do make mistakes! He should have multiplied by .75 instead of dividing.

3. Mar 8, 2009

### lanedance

I think you make a fair point & I agree with your way

the energy absorbed by the water should not be more then the energy output...