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Time required for a liquid to change in Temp (thermo)

  1. Feb 9, 2010 #1
    Hi this is my first post here.

    I am attempting to calculate how long it takes for a soda can(355ml) to change from Current room temperature(75F), to say 38F, however the refrigerator or ambient temperature in this case is 35F.

    I was given a corollary from a friend, but the values I get in return are way off.

    t = pV/hA * LN[ (T1 - Ta)/(T2 - Ta) ]

    t= [(988.9 kg/m3 )(0.000355 m3 )]/[(255W/m2K)(.0277m2)] * LN[ (75 - 35)/(38 - 35)]

    I attempted to use metric for all the units.
    p=density
    V=Volume
    A=surface area
    h= convection coefficient of water or the aluminum can?
    T1 and T2 are the initial and final temperatures of the drink.
    Ta is the ambient temperature.

    However I plug-in the equation it seems to not output realistic data. I'll get something like .5 seconds depending on what variables I plug in/manipulate.

    Is my equation incorrect or my units incorrect? Also are their any other better equations to do this with? This is driving me mad, please help.

    Ideally I need to figure out a working base equation, as well as know what each unit is.
     
  2. jcsd
  3. Feb 9, 2010 #2

    Mapes

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    Hi movetwice, welcome to PF. Your equation has to be wrong, because it gives time in units of kg W K-1!

    Presumably your friend is assuming the temperature of the liquid in the can is uniform, and the heat lost equals the heat transferred by convection:

    [tex]\rho Vc\frac{dT}{dt}=hA(T-T_\infty)[/tex]

    which is solved to give

    [tex]t=\frac{\rho Vc}{hA}\ln\frac{T_i-T_\infty}{T-T_\infty}[/tex]

    And the convection coefficient looks way too high. I would think 25-50 W m-2 K-1 would be more reasonable, but others may want to weigh in with their estimates.
     
  4. Feb 9, 2010 #3
    excellent thank you!, this might be a dumb follow up question, but what is the unit for "c"?
     
  5. Feb 9, 2010 #4

    Mapes

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    [itex]c[/itex] is the http://en.wikipedia.org/wiki/Specific_heat" [Broken].
     
    Last edited by a moderator: May 4, 2017
  6. Feb 9, 2010 #5
    thanks!
     
  7. Feb 10, 2010 #6
    Alright so I'm still running into issues with this equation. I'm thinking my units are off somehow, because allegedly a can of soda is supposed to get to optimal temperature in 20-25 minutes in a freezer.

    t = pVc/hA * LN[ (T1 - Ta)/(T2 - Ta) ]
    t= [(988.9 kg/m3 )(0.000355 m3 )(4.186)]/[(25W/m2K)(.000277m2)] * LN[ (75 - 0)/(38 - 0)]

    t= 2.12207 * 0.679901 = 1.44279

    p = 998.9 kg/m3 (standard density of water)
    V= 355ml = .000355 m3
    C= standard C of water 4.186 J/g*C
    h= 25 W/m2k
    A= 277 cm^2 = .0277 m^2 (or sqm)

    Any advice on where im going wrong? I'm looking to get a working equation that will be highly accurate so I can isolate and change certain variables and get real usable time calculations.
     
  8. Feb 10, 2010 #7

    Mapes

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    Have you checked your units? At the end, you should be left with seconds.
     
  9. Feb 12, 2010 #8

    GT1

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    H for free convection is usually not more then 5-7 W/m^2*K
    Also the temperature should be in K or degC
     
    Last edited: Feb 12, 2010
  10. Feb 12, 2010 #9
    For consistent units, you need c in J/kg*C, which is about 4190.
    This will change the result by a factor of 1000.
     
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