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Specific heat capacity (Calorimetry)

  1. Feb 10, 2008 #1
    [SOLVED] Specific heat capacity (Calorimetry)

    1. The problem statement, all variables and given/known data
    When a driver brakes an automobile, friction between the brake disks and the brake pads converts part of the car's translational kinetic energy to internal energy. If a 1500 kg automobile traveling at 32 m/s comes to a halt after its brakes are applied, how much can the temperature rise in each of the four 3.5 kg steel brake disks? Assume the disks are made of iron [tex](c_{p} = [/tex] 448 J/kg•°C) and that all of the kinetic energy is distributed in equal parts to the internal energy of the brakes.


    2. Relevant equations

    [tex]c_{p,x}m_{x}T_{x}[/tex] = [tex]c_{p,y} m_{y} T_{y}[/tex]




    3. The attempt at a solution
    I really don't know where to start with this problem... A little push and I can do the rest, but...

    [tex]c_{p,x}m_{x}T_{x}[/tex] = [tex]c_{p,y} m_{y} T_{y}[/tex]
     
    Last edited: Feb 10, 2008
  2. jcsd
  3. Feb 11, 2008 #2
    Bump. Can anyone help me out?
     
  4. Feb 11, 2008 #3
    Well that's not really the right equation involving specific heat you're looking at

    You want the one that involves energy. You know the kinetic energy of the car from 1/2*mv^2, and it's given that all that energy becomes heat

    A 4th of that goes into each 3.5kg brake, so do you have a formula that links specific heat, mass, temperature, and heat? (yes you do)
     
    Last edited: Feb 11, 2008
  5. Feb 11, 2008 #4
    [tex]\Delta[/tex]PE + [tex]\Delta[/tex]KE + [tex]\Delta[/tex]U = 0?
     
  6. Feb 11, 2008 #5
    Well you could infer the equation from the units if you didn't know off hand(energy/mass*temp

    c=Q/(m*dT) where dT is the change in temperature and Q is the amount of heat gained(which you find from assuming all the kinetic energy becomes heat)
     
  7. Feb 11, 2008 #6
    Thanks. I got it. It's 120 degrees Celsius, right?
     
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