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Calorimetry specific heat

  1. Oct 16, 2014 #1
    1. The problem statement, all variables and given/known data
    A 0.080kg copper container (specific heat: 387Jkg-1K-1) contains 0.30kg of water and 0.040kg of ice at 0°C. Steam at 100°C is passed into the water and its temperature stabilizes at 20.0°C. Find the mass of the water left in the container assuming the system is insulated.


    2. Relevant equations
    Q= mcΔT
    Q= mL


    3. The attempt at a solution
    Q lost = Q gained
    msteamcΔT + msteamL = mCucΔT + mH20cΔT + micecΔT + miceL
    msteam = mCucΔT + mH20cΔT + micecΔT + miceL / cΔTsteam + Lsteam
    Total mass water= mass steam + water + ice

    Is this correct?
    Is ΔT for steam -80 (20-100°C)?
     
  2. jcsd
  3. Oct 16, 2014 #2

    ehild

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    Homework Helper
    Gold Member

    Welcome to PF!

    Do not use the same symbol ΔT for different things. The change of temperature of water is not the same as the change of temperature of the deposited steam.
    Your equation means that the heat supplied by the steam is equal to the heat absorbed by the colder parts of the system.
    It is right that the change of temperature is -80° for the deposited steam, but you have to use 80° when you calculate the supplied heat.
    The equation for msteam is not correct without parentheses around cΔTsteam + Lsteam

    ehild
     
  4. Oct 16, 2014 #3
    msteam = mCucΔTCu + mH20cΔTH20 + micecΔTice + miceL / (cΔTsteam + Lsteam)

    Is this right?
     
  5. Oct 16, 2014 #4

    ehild

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    You need parentheses also around the nominator. And be sure using correct sign in the denominator.

    ehild
     
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