# Refrigerator question

1. Sep 11, 2004

### Ylle

Well I'm new, and I have a question :)

I've been looking and searching lots of places, but no luck :(
But my problem, which isn't really a problem, it's just something I would like to know :)
I was wondering if there were a formula, that could find out how long time a glass of water or a beer can for example, that got a temperature of 20 degrees Celsius, would have to be in a refrigerator, that got a temperature of 5 degrees Celsius (For example), for it to get the temperature 7 degrees Celsius for example.

I don't know if there is an easy formula for this, but I hope so, and I hope you understand what I mean.

- Ylle

2. Sep 11, 2004

### Tide

You could use Newton's Law of Cooling which says that the rate of cooling is proportional to the temperature difference between the object and the ambient temperature:

$$\frac {dT}{dt} = -k (T-T_A)[/itex] Of course, you'll need a couple of data points to determine the value of the proportionality constant k. 3. Sep 11, 2004 ### PRodQuanta d=delta (change in) T=temp t=time [tex]T_A$$=ambiant temp.

Just to let you know if you were confused. If you already knew this, sorry for insulting your intelligence. Wasn't meant for that.

4. Sep 11, 2004

### Integral

Staff Emeritus
You could apply Newton's Law of Cooling to this problem.

$$\frac {dT} {dt} = -k (T-T_r)$$

where T is the temperature of the beer and Tr is the temperature of the refrigerator. The solution to this is

$$T(t) =c e^{-kt}+T_r$$

The constants c and k need to be determined. For c let Ti= initial temperature of the beer we get:

$$T(0)=T_i = c + T_r$$ so
$$c = T_i-T_r$$

To get k you will need to find the temperature of the beer after some time has passed, say 10 min. Let T(10) be the temperature after 10 min, solving for k we get.

$$k = - \frac {ln \frac {T(10) - T_r} c} {10}$$
now combine the constants and the original solution to find the temperature at any time.

5. Sep 11, 2004

### LURCH

I see nothing in these equations to account for the ratio of volume to surface area. Does this method assume a spherical object? And, would the ratio have negligable effects on beer cans and glasses of soda?

6. Sep 11, 2004

### Integral

Staff Emeritus
That is why the k needs to be determined by experiment. It will be different for different containers. It will even be effected by near by objects. Say if you place a large block of Ice beside the can vs a pot of boiling water. Or if the can is in a 6 pack vs alone on the shelf. This is indeed a rather simplistic model but such a problem gets vastly difficult in a hurry if you attempt to get detailed.

The k is a bulk parameter determined for a specific situation.

Last edited: Sep 11, 2004
7. Sep 12, 2004

### Ylle

What did you say Ti was. I'm danish, so I don't know so many english physical words. But is it like: C =m*c where c is the c i need (Heatcapacity if you translate it directly to danish), or is it another c ?
And does k need to be determined for every different time you have ?
Because you say: $$k = - \frac {ln \frac {T(10) - T_r} c} {10}$$ where T(10) is T(t) for 10 minutes: $$T(10) =c e^{-k10}+T_r$$ or is a constant you can use every time ?

Well, I'm kinda wasted right now, so I excuse my english mistakes :)

8. Sep 12, 2004

### Integral

Staff Emeritus
Ti is the starting temperature of the beer.

I would say that you need to determine a k for each different type of container or if it is an individual can or a 6 pack. All of the physical constants related to the beer and container are "bulked" together into this single bulk constant which is determined by experiment.

Once you find a k for a particular container and beverage (the type of beer could effect it, say American P*ss vs Guinness) you should be able to make pretty good estimates on the time required to reach drinking temperature.

Last edited: Sep 12, 2004
9. Sep 12, 2004

### Ylle

Hmmm

I was wondering...
You are writing:
$$T(t) =c e^{-kt}+T_r$$

And then you are writing:
$$k = - \frac {ln \frac {T(10) - T_r} c} {10}$$

But how am I suppose to figure out k, when I need T(t) (T(19)) in the formula ? And then I need k in the T(t) formula, so basicly it's an evil circel isn't it

10. Sep 12, 2004

### Integral

Staff Emeritus
T(19)? .... I'll assume that you mean T(10), this needs to be MEASURED with a thermometer after the beer has been in the refrigerator 10 minutes. The 10 in the denominator is the 10min time. If you measure after 5 min use a 5 in the denominator.