# Heat transfer

1. Nov 15, 2007

### sphynx

Not sure whether anyone can help, I need to work out the rate at which butter reaches ambient room temperature from being chilled at various temperatures.

Can anyone point me in the right direction?

2. Nov 15, 2007

### stewartcs

3. Nov 15, 2007

### sphynx

Thanks

4. Nov 15, 2007

### katchum

You'll need the k coefficient of butter and the convection coefficient of the system of the surface of the butter and the sides of the butter.

Try making a series of resistances of: butter conduction + parallel sides of convection of the butter + upper surface convection. For butter conduction I would suggest using a spherical equation. Your Rconduction will be function of the diameter of the butter and the k coefficient. For the others you use convection over a vertical wall and convection over a horizontal plate.

If you have the series of resistances R you can then say that Q is the flow of warmth, which is the same for every part of the butter (conduction and convection).

In this case you can say that there is a constant room temperature and you have an initial temperature of the upper surface of the butter. Q = h (Tr-Ts) and h depends on what kind of surface you have, you'll have to use correlations to find this one. Tr is constant.

When you have the Q, you can make little time intervals, use a numerical program, or you can make yourself (in Matlab) a loop with a certain time interval deltat. Q will constantly change and Ts will change too.

Now why did we need this R then? It's just: when you know Q and you know R, you will know delta Temperature between inside the butter and the outside ambient temperature.

So you will have Tbutter in function of temperature because: R = deltaT/Q

I'm afraid you can't solve this one in an easier way, even when you only want the surface temperature of the butter, you'll still need to take in account the conduction between surface and inner butter. I'm also afraid you'll need to consider convection because it's this 'natural' convection that will melt the butter.

The Newton thing is a bit simplistic, because you don't take in account that the butter has a gradient and convection has not been taken account of. But it will do...

Last edited: Nov 15, 2007