How Does Specific Heat Relate to Force in Stopping a Car?

[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.

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



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

 

[wais]

The specific heat, Q, is a measure of the amount of heat energy required to raise the temperature of a substance by a certain amount. In this problem, the brakes of the car are cooling off, which means that heat energy is being transferred from the brakes to the surrounding air. The force, F, is related to this process because it is the force that is being applied to the brakes to stop the car.

To calculate the total entropy change in this scenario, we need to consider the change in temperature of the brakes and the surrounding air. The specific heat of the brakes and the surrounding air will determine the amount of heat energy that is transferred and ultimately the change in entropy.

To relate specific heat and force, we can consider the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. In this case, the force applied to the brakes causes a decrease in the car's kinetic energy, resulting in a decrease in temperature and a transfer of heat energy. The specific heat of the brakes determines the amount of heat energy required for this temperature change.

In terms of units, the unit for specific heat is J/kg·K, which means that it takes 1 joule of energy to raise the temperature of 1 kilogram of a substance by 1 Kelvin. The unit for force is N, which is equivalent to kg·m/s^2. So, in this scenario, we can say that the force applied to the brakes results in a change in temperature, which in turn affects the specific heat and ultimately the total entropy change.