Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Refrigerator question

  1. Sep 11, 2004 #1
    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. jcsd
  3. Sep 11, 2004 #2

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    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:

    [tex]\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.
     
  4. Sep 11, 2004 #3
    d=delta (change in) T=temp
    t=time
    [tex]T_A[/tex]=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.

    Paden Roder
     
  5. Sep 11, 2004 #4

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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


    [tex] \frac {dT} {dt} = -k (T-T_r)[/tex]

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

    [tex] T(t) =c e^{-kt}+T_r[/tex]

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

    [tex]T(0)=T_i = c + T_r [/tex] so
    [tex]c = T_i-T_r [/tex]

    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.

    [tex] k = - \frac {ln \frac {T(10) - T_r} c} {10} [/tex]
    now combine the constants and the original solution to find the temperature at any time.
     
  6. Sep 11, 2004 #5

    LURCH

    User Avatar
    Science Advisor

    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?
     
  7. Sep 11, 2004 #6

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
  8. Sep 12, 2004 #7
    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: [tex] k = - \frac {ln \frac {T(10) - T_r} c} {10} [/tex] where T(10) is T(t) for 10 minutes: [tex] T(10) =c e^{-k10}+T_r[/tex] or is a constant you can use every time ?

    Well, I'm kinda wasted right now, so I excuse my english mistakes :)
     
  9. Sep 12, 2004 #8

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
  10. Sep 12, 2004 #9
    Hmmm

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

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

    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 :confused:
     
  11. Sep 12, 2004 #10

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Refrigerator question
  1. Refrigerator Magnet (Replies: 6)

Loading...