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Math Help - [SOLVED] Chem 1000: Specific Heat problem

  1. #1
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    [SOLVED] Chem 1000: Specific Heat problem

    Folks,
    Im not understanding how to find the specific heat of something. The books examples are a bit confusing! Could someone please break it down for me?

    Here's the problem:
    Assuming that Coca-Cola has the same specific heat as water, how much energy in calories is removed when 350g of coke is cooled from room temp (25 degrees C) to refrigerator temp (3*C)?
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  2. #2
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    Specific heat is the amount of energy needed to heat 1 gram of a substance by 1 C. The specific heat of water is 1 cal/g*C. That means, if you give, say, 1 cal as heat to 10 grams of water, its temperature will increase by 0.1 C.
    You should have see this formula in your physics/chemistry class:

    Heat = m*c*\Delta T

    where m is the mass in g, c is the specific heat, and \Delta T is the temperature differential in C.

    Replace the variables in the formula and you have the solution to your problem (remember, the temperature differential is negative, since the water looses heat!).
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  3. #3
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    e^(i*pi)'s Avatar
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    To add to Referos answer the specific heat of water is 4.18 kJ/(kg K) if you're using SI units
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  4. #4
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    So is this correct?
    3.2186x10^4
    I used Referos formula and the specific heat that e^(i*pi) gave me:
    350*4.18*22
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  5. #5
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    Quote Originally Posted by ninobrn99 View Post
    So is this correct?
    3.2186x10^4
    I used Referos formula and the specific heat that e^(i*pi) gave me:
    350*4.18*22
    Your answer is correct if your result is in J

    To get an answer in kJ divide by 1000 or convert the 350g into kg by dividing by 1000 = 0.35kg

    0.35*4.18*-22 = -32.186kJ. It will be negative because heat is lost
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  6. #6
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    haha, I was just about to post that I mixed the two units up. Thank you again.
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