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Homework Help: What is the heat capacity of the tires of the car?

  1. Jan 26, 2009 #1
    1. Assume that a car who has a mass of 1000kg (including driver) is initially rolling freely (without using the engine) down a 30° slope at a speed of 5m/s. The brakes are used causing the car to stop after 10 m.

    Find the HEAT CAPACITY of the brakes if they rise in temperature by 10 K.

    Drag forces are negligible, no heat transfer to or from the car occurs and the only change in U (internal energy) is due to heating of the brakes

    2. I need to find Q

    Q = mc(Tf-Ti)

    Using conservation of energy

    W + Q = ΔK.E + ΔP.E + ΔU

    W = sin(30) * d

    K.E = (1/2)mv2

    The rest I do not know?!

    3. The answer is C =6.25 x 103 but I do not know how this is obtained.

    What confuses me is the part that says that ΔU is not zero and that there is no heat transfer to or from the car.

    What is the correct method in order to obtain that answer?

    Thank you
  2. jcsd
  3. Jan 26, 2009 #2


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    Q = mc(Tf-Ti) is specific heat capacity, since you are not told the mass of the brakes - all you can do is work out the temperature change for the input energy.

    You have Q = C(Tf-Ti), you have T, you found Q from the loss in ke and pe of the car, just drop it into the equation to get C

    Thats just to make the problem solvable - you don't have to calculate the cooling of the brakes
  4. Jan 26, 2009 #3
    How do I find the change in potential energy though?

    Wouldnt I also need to know work and internal energy in order to find Q?

    I do not understand...could you be kind enough ( or anybody else) to explain more?
  5. Jan 26, 2009 #4


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    Draw a diagram, the car goes 10m along the hypotonuse of a triangle an angle of 30deg - so you can work out how far vertically this is. together with the mass this gives you change in PE.
    You are given the initial speed and presumably it brakes to zero, so you have the change in KE.

    Thats the total energy going into the brakes, the heat capacity is just how many Joule/kelvin they rise in temperature
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