Calculating Change in Entropy: 1100 kg Cars Colliding at 75km/hr

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    Entropy Homework
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To estimate the change in entropy from the collision of two 1100 kg cars traveling at 75 km/hr, one can use the equation ∆S = k ln(W) where k is the Boltzmann constant and W represents the number of microstates. The collision is inelastic, resulting in kinetic energy being converted into heat, which contributes to the entropy change. The heat (Q) generated can be calculated from the cars' mass and speed using the kinetic energy formula. It's essential to define the system's boundary to analyze the entropy change accurately. Understanding these principles will facilitate the calculation of entropy for this scenario.
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Homework Statement

Two 1100 kg cars are traveling 75 km/hr in opposite directions when they collide and are brought to rest. Estimate the change in entropy of the universe as a result of this collision. Assume T= 15oC



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The Attempt at a Solution

PLEASE GET ME STARTED I am physics retarded an equation would be nice...
 
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i imagine you're meant to use the relationship ∆S=k*log(w) where ∆s=entropy change, k is 1.380 6504(24)×10-23 J/K, and w is work...
work is kinetic energy, so...
got it?
 


Entropy can be calculated as:


S = k ln W ( w not as in work)

also you can use S = Q/T ---> Q being heat T ----> temperature,
I have learned that in thermo when they refer at universe is just whatever is outside your system. so define your boundary and i think it's pretty simple.
 


Also forgot to mention... you have an inelastic collision where two masses have momentum.! Remember that there is a release of energy in the form of heat because of the collision. I think that's the key for you to plug the formula as mentioned above.

However that guy who said w is work WRONG WRONG WRONG

W is the Wahrscheinlichkeit, the frequency of occurrence of a macrostate or, more precisely, the number of possible microstates corresponding to the macroscopic state of a system — number of (unobservable) "ways" in the (observable) thermodynamic state of a system can be realized by assigning different positions and momenta to the various molecules.

I know this because I just did a test on classical thermodynamics at uni ! =)
hope this helps you
 


Wait, how do I get Q (heat) from my mass and speed?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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