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Specific heat, Q, and force

  1. Oct 23, 2004 #1
    problem: a 1350 kg car is moving at 23m/s. the driver brakes to a stop. the brakes cool off to the temperature of the surrounding air, which is nearly constant at 15 celsius, what is the total entropy change?

    i know that S=integral of dQ/T since T is constant i can take it out of the integral and integrating dQ is just Q so s= Q/T

    how can specific heat, Q, relate with force F

    i konw that the units for Q is J , and F is N (which is kg*s^2)

    is it simply just converting the units from F to J? if so, what are the units for J?
  2. jcsd
  3. Oct 23, 2004 #2


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    Use Conservation of Energy.

    [tex] Q_{dissipated}=m\frac{v^2}{2}[/tex] both quantities in Joules.


    [tex] \Delta S=\frac{Q_{dissipated}}{T}[/tex] in Joules /Kelvin
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