specific heat, Q, and force

[nemzy]

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?

[Clausius2]

Use Conservation of Energy.

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

:smile:

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