How Does Stirring Affect Water's Temperature Increase?

[cowmoo32]

Here's my question:
Suppose you warm up 550 grams of water (about half a liter, or about a pint) on a stove, and while this is happening, you also stir the water with a beater, doing 4e4 J of work on the water. After the large-scale motion of the water has dissipated away, the temperature of the water is observed to have risen from 24°C to 75°C.
What was the change in the thermal energy of the water?
The answer is 1.18e5 Joules

The equation is Delta E = W + Q = mC*(Delta Temp)
Solve for Q m=550 C=4.7 change in temp=51 and I'm guessing work is 4e4, but solving for Q using those numbers gives me 2.945 Joules. What am I messing up?

 

[OlderDan]

It sounds to me like you are misinterpreting the problem. Are you really being asked to find Q?

 

[cowmoo32]

yeah, it asks for the change in thermal engergy of the water and the only unkown I have is Q..

Here's the solution my online homework gave me:
http://img85.imageshack.us/img85/3900/screengi2.jpg

 

[OlderDan]

yeah, it asks for the change in thermal engergy of the water and the only unkown I have is Q..

Here's the solution my online homework gave me:
http://img85.imageshack.us/img85/3900/screengi2.jpg

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What does thermal energy of the water mean?

 

[cowmoo32]

well, Delta E thermal for the water is change in the heat engergy of the water, which is Q, is it not?

 

[cowmoo32]

wait...i can't believe i didnt get this.

DeltaE = mC*DeltaT, and I have all of those
I still get 117810 and the answer is 118000..

 

[OlderDan]

well, Delta E thermal for the water is change in the heat engergy of the water, which is Q, is it not?

In this case ΔE is not Q. The thermal energy of the water is increased by the heat added Q plus the work done by the beater W. That is what your online slolution/explanation is saying. But you are not being asked for Q, you are being asked for ΔE.