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Calorimetry: Thermal energy going into translational KE

  1. Jan 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose a cup of boiling water (m=250g) instantaneously cools to room temperature (25°C) with the liberated thermal energy going into translational KE. How fast will the cup fly off the table? Assume the water molecules have 18 degrees of freedom.

    2. Relevant equations

    Q=mcΔT
    Uwater=9nRT (not sure about this one)


    3. The attempt at a solution

    So what I first did was try to calculate Q:

    Q=mcΔT
    Q=(250g)(1)(25-100)
    Q=-18750 cal

    I then converted calories to joules:

    -18750 cal x 4.186 J = -78487.5 J.

    I am not sure where to go after this. Any help?
     
  2. jcsd
  3. Jan 22, 2012 #2
    Update:

    I'm guessing that since the cup is on a horizontal table, the degrees of freedom is 9. Therefore the energy must be halved.

    Q = 39243.75 J

    Plug it into:

    KE = (1/2)mv2

    v2 = (2KE) / m
    = (2*39243.75) / .250
    = 313950

    √v2 = √313950
    v = 560 m/s

    Can anyone confirm that this is correct?
     
  4. Jan 24, 2012 #3
    Any help?
     
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