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Thermal physics need help.

  1. May 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Dry steam is used to make a cup of coffee by bubbling it through water. If the
    steam is at 100°C, what mass of steam must be used to heat 200 g of water from
    25°C to 95°C?


    2. Relevant equations
    Not quite sure but I think:
    Q= mCΔT
    Qcold=-Qhot


    3. The attempt at a solution

    Ok so I started out by trying to use Qcold=-Qhot, and this was what I did:

    Hot:
    m=??
    C=2020 (this is what the textbook said the specific heat of steam was)
    Ti=100°C
    Tf= 95°C (because I assumed that they would end up the same temp because of equilibrium)

    Cold:
    m=0.2kg
    C=4200 (textbook said this was specific heatr for water)
    Ti=25°C
    Tf=95°C

    So then I subbed in values to come up with:

    4200x0.2(95-25)=-(2020m(95-100))
    58800=10100m
    m=5.82kg

    this is a ridiculous number and the text book says that the answer is 26g.

    Please help me understand this.
     
  2. jcsd
  3. May 17, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    Think "phase change" :wink:
     
  4. May 17, 2012 #3
    You left out the heat of condensation of the steam.
     
  5. May 17, 2012 #4
    yeah thanks guys I realised just then the answer should have been 58800=2270100m, m= 58800/2270100= 0.0259kg=25.9g=26g. Thanks for that guys I usually do really well in physics so when i can't understand something I start to stress ahah. Much love and God Bless
     
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