# Thermal Physics and Bottle Shake

1. Jan 22, 2012

### sportcardinal

1. The problem statement, all variables and given/known data
If you were to shake bottle with some water in it for about ten minutes, then roughly, how much would the temperature increase?

2. Relevant equations

K=1/2mv^2
U=N*f*1/2*k*T

3. The attempt at a solution

I think we need to find kinetic energy and assume it all gets converted to thermal. But I don't understand how to get temperature.

Do I just find K, then assume that is U and solve for T?

2. Jan 23, 2012

### rude man

Depends on whether your bottle is an insulator, like a Thermos, or whether it's conducting to the ambient, or something inbetween (which of course will be the case).

Assuming a perfect insulator, and ignoring the heat capacity of the bottle itself (imagine a very thin, perfectly insulating bottle), all the work W you do in shaking the water is tranferred to internal energy buildup of the water. You can estimate the resulting temperature buildup from dW/dT = CV where CV is heat capacity of the water (= specific heat of water, 1 cal/gm, times no. of grams m in your bottle, or CV = mCv. So for every erg of shaking you'd get 1 deg C temp. buildup divided by the no. of grams of water: ΔT = W/CV.

If you allow heat to escape from the bottle as you shake it, that of course changes the picture. If a perfect conductor there would be no change in the water temp.

3. Jan 28, 2012

### LawrenceC

You are also assuming there is no change in internal energy of the air in the bottle else that would have to be included, however slight. In the problem statement the word 'some' implies the bottle is not entirely full of water.

4. Jan 28, 2012

### rude man

Yes, but, as you say, that's very slight. C of water = 1 cal/cc-K;
C of air at 1 at. and 20C = 0.29e-3 cal/cc-K.

5. Jan 28, 2012

### LawrenceC

"imagine a very thin, perfectly insulating bottle"

If you make this assumption, you must make the other.