Entropy of Macroscopic Collision

In summary, the problem involves two cars colliding head-on at 91 km/h, with a mass of 1410 kg each. The total entropy change of the cars and environment system due to the collision is calculated by adding the changes in entropy of the cars and the environment. This can be related to the formulas and equations in the textbook by considering the conversion of kinetic energy into heat energy, as well as the work done by non-conservative forces. Ultimately, the assumption is made that the collision is an isothermal process, resulting in a delta-U and delta-T of 0, and a delta-S of Q/T.
  • #1
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Homework Statement


Two cars each of mass 1410 kg traveling at 91 km/h in opposite directions collide head-on and come to a disastrous halt; see Figure P.38. Assume that the surrounding air and ground temperature remains fixed at 18°C. Calculate the total entropy change of the cars and environment system that results from the collision.



Homework Equations





The Attempt at a Solution


I understand how to do most problems involving gases or mixing liquids or engines etc. which involve usually microscopic untangible processes but I'm not sure what to make of this. As soon I see collision I think momentum but I don't know where to begin to relate collisions to the formulas and equations in my textbook which deal with volumes, pressures, and mols to this. any hint as a jumpstart would be appreciated! thank you
 
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  • #2
S(total)=S(car) + S(car 2) + S(environment) where the values are changes in entropy

delta-S=delta-q/T, where delta-q=heat and T is a constant temperature?

how do I relate velocities and masses to heat?
 
  • #3
I am really just guessing here, but perhaps all of that kinetic energy is assumed to be converted to heat energy ?
 
  • #4
wait since
Q + W'=delta-U, where W'=work done by nonconservative/other forces i.e. the work done by the surroundings on the system

W'=(delta-KE) + (delta-PE)...in this case PEi=0 and PEf=0
thus W'=1/2mvf^2- 1/2mvi^2 for both cars

I know just intuivitely that when a car crashes the KE of both cars is translated into heat energy and some into sound and even light (which revert back to heat)...
 
  • #5
oh haha ok maybe that's the assumption to make because otherwise calculating the delta-U would be difficult or is the delta-U of the entire system=0 i.e. unchanged?
 
  • #6
well wouldn't it be? it's an isothermal process?
 
  • #7
uhh nevermind I got it
took a little reasoning but not too difficult
isothermal so delta-T=0 i.e. delta-U=0
Q + W'=U-->Q=-W'
W'=(delta-KE)
delta-S= Q/T

thanks
 

Related to Entropy of Macroscopic Collision

What is entropy?

Entropy is a measure of the disorder or randomness in a system. It is a thermodynamic property that describes the distribution of energy within a system.

What is the entropy of a macroscopic collision?

The entropy of a macroscopic collision is a measure of the increase in randomness or disorder in the system after the collision occurs. It takes into account the change in energy distribution and arrangement of particles.

How is entropy related to the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system will always increase over time. This means that in a macroscopic collision, the entropy of the system will increase due to the randomization of energy and particles.

What factors affect the entropy of a macroscopic collision?

The entropy of a macroscopic collision is affected by several factors, including the types of particles involved, their initial energy distribution, and the nature of the collision (elastic or inelastic).

How is the change in entropy calculated for a macroscopic collision?

The change in entropy for a macroscopic collision can be calculated using the formula ΔS = Q/T, where ΔS is the change in entropy, Q is the heat transferred during the collision, and T is the temperature of the system. Alternatively, it can be calculated by comparing the entropy of the system before and after the collision.

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