Two cars,each of mass 2 200 kg and traveling at 14.0 m/s,undergo a head-on completely inelastic collision.What is the change in entropy of the Universe which results from this collision? Assume the ambient temperature is 20.0 $\circ$ C.

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Quizplus · Accepted Answer

The change in entropy of the Universe resulting from this collision can be calculated by first determining the kinetic energy lost during the collision, which then gets converted into heat. Since the collision is completely inelastic, the two cars stick together and their final velocity is zero, meaning all kinetic energy is converted into heat.The initial kinetic energy (KE) of each car is given by $KE = \frac{1}{2}mv^2$, where $m = 2200 \, \text{kg}$ and $v = 14.0 \, \text{m/s}$. Therefore, the total initial kinetic energy of both cars is $2 \times \frac{1}{2} \times 2200 \times 14.0^2 = 431200 \, \text{J}$.This energy is then converted into heat, $Q = 431200 \, \text{J}$, and the change in entropy ($\Delta S$) can be calculated using the formula $\Delta S = \frac{Q}{T}$, where $T$ is the absolute temperature in Kelvin. Given the ambient temperature is $20.0^\circ\text{C}$, which is $293.15 \, \text{K}$ (since $20.0^\circ\text{C} = 293.15 \, \text{K}$), the change in entropy is:$$\Delta S = \frac{431200}{293.15} \approx 1470 \, \text{J/K} = 1.47E+3 \, \text{J/K}$$Therefore, the correct answer is A) 1.47E+3 J/K.

Two Cars,each of Mass 2 200 Kg and Traveling at 14.0