# Thermodynamics, Heat Transfer, differential equals

1. Oct 10, 2009

### al3ko

Hi there,
first of all:
This is not a homework task or something like that. It is just one of my interests and I want you to check my calculation, if my thoughts are correct or not.
Furthermore I apologize for my incorrect englisch but I am a foreigner.

Okay, now some background information:
Imagine a waterreservoir in the size of a bathtub and a constant watertemperature. You put a beerbottle into that reservoir and you want to describe the changing of the temperature of the beer. Allright. In my opinion, we have two physical phenomens:
1. Heat transfer from the reservoir to the bottle
2. Heat transfer from the bottle to the beer.

Is it right?
Okay, so the first equation should explain the incoming heat energy from the reservoir and the outgoing heat energy to the beer connected with the energy changing of the bottle.
$$$$\dot{Q}_{res}-\dot{Q}_{beer}=\frac{dU_{botl}}{dt}$$$$
The second equation bescribes the inner energy of the fluid by incoming heat energy from the bottle:
$$$$\dot{Q_{beer}}=\frac{dU_{beer}}{dt}$$$$

I just want to know: Is there any mistake in the energy bilances? This is the basic I have to go on working with. If this is wrong everything else is wrong, too :)

Therefore I just want you to give a short response if my thoughts are right or not.

Best regards
al3ko

2. Oct 10, 2009

### Franco_v

It looks correct to me.

3. Oct 11, 2009

### al3ko

Hi franco,
thanks for your statement.

I was just a little bit unsure about the first equation with the outgoing power transfer from the bottle into the liquid. [tex]$$...-\dot{Q}_{beer}=...$$[\tex] because that complicates my solving of the ODE so I think I have to do it numerical by Matlab.

Best regards
al3ko