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Homework Help: Heat problem

  1. Feb 6, 2008 #1
    [SOLVED] Heat problem

    1. The problem statement, all variables and given/known data
    The manager of a fried chicken place wishes to store a ton of poultry at 0 degrees Fahrenheit. Initially the poultry is at 20 degrees Celsius. Calculate the energy in BTUs that is removed in this cooling for storage process.

    Assume that the poultry freezes at 32F

    Latent heat: 99 BTU/lb
    Specific heats: 0.80 BTU/deg F/lb (above freezing)
    0.41 BTU/deg F/lb (below freezing)

    2. Relevant equations
    Was not specifically given any but from the book I'm thinking...

    Specific heat
    Q = cm[tex]\Delta[/tex]T

    Latent heat
    L = [tex]\frac{|Q|}{m}[/tex]

    3. The attempt at a solution
    Would I even have to use any one of those equations? I am already given the specific heat and latent heat so those should do me no good right?

    [tex]\Delta[/tex]T = 20 - 0 = 20

    Q = (.8)(2000)(20)
    Q = 32000J

    I know I should have to convert all fahrenheit to celcius.

    I am really at a loss though what I can use to give me a final BTU. If anyone could give me a push I'd appreciate it. I have no other sources but Internet and the book so other books will not help at this point. Thanks.
  2. jcsd
  3. Feb 6, 2008 #2

    Shooting Star

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    Homework Helper

    Then why don't you? Stick to some consistent set of units.

    In the eqn Q = (.8)(2000)(20), where is the 2000 coming from? The 0.8 is in BTU/lb/F, but delta_t = 20...This is meaningless!

    Change everything to one system of units.
  4. Feb 6, 2008 #3
    Whoops...sorry I meant to put in what each letter stood for.

    Q = cm[tex]\Delta[/tex]T
    [tex]\Delta[/tex]T - change in temp
    m - mass
    c - specific heat (which is given)

    1 ton = 2000 lbs = 907.184 kg
    kW = Btu / 3414

    Q = (.8 BTU/lb/F)(907.184kg)(20C)
    Q =

    Can I convert the BTUs to kW just by using the equation kW = BTU/3414?
  5. Feb 6, 2008 #4

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    1 ton = 2240 lb = 907 kg.

    As I said, convert everything to one system of units. Sp heat is given in BTU. Temp change in C can be easily converted to F. Use whichever you like.

    > Can I convert the BTUs to kW just by using the equation kW = BTU/3414?

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