# Thermal Engergy and Water

1. Nov 13, 2006

### 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 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?

2. Nov 13, 2006

### OlderDan

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

3. Nov 13, 2006

### 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 hw gave me:

4. Nov 13, 2006

### OlderDan

What does thermal energy of the water mean?

5. Nov 13, 2006

### cowmoo32

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

6. Nov 13, 2006

### 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..

7. Nov 13, 2006

### OlderDan

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.