Two 1100 kg cars are traveling at a speed of\({\bf{85 km/h}}\)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\(T = {\bf{20\circ C}}\).

The two cars collide with each other and then come to rest. The first car is moving with some kinetic energy.

The energy lost during a collision is equal to the total kinetic energy. Both have the same mass and the same velocity.

The total energy lost during a collision is two times the kinetic energy.

The given data can be listed as

• The mass of each car is\(m = 1100{\rm{ kg}}\).
• The velocity of each car is\(v = 85{\rm{ km/h}}\left( {\frac{{1{\rm{ m/s}}}}{{3.6{\rm{ km/h}}}}} \right) = 23.61{\rm{ m/s}}\).
• The temperature value is\(T = 20\circ {\rm{C}} = \left( {20\circ {\rm{C}} + {\rm{273}}} \right){\rm{ K}} = 293{\rm{ K}}\).

During the collision, the kinetic energy of the cars is lost. The total kinetic energy becomes unusable after the collision. So, this kinetic energy is transferred to the surrounding environment.

The surrounding of the universe utilizes this energy. This will result in a change in the entropy of the universe.

The total kinetic energy can be expressed as

\(\begin{array}{c}{K_{\rm{T}}} = {K_1} + {K_2}\\ = \frac{1}{2}m{v2} + \frac{1}{2}m{v2}\\ = m{v2}\end{array}\).

Here,\({K_1}\)is the kinetic energy of the first car,\({K_2}\)is the kinetic energy of the second car,\(m\)is the mass of the car, and\(v\)is the velocity of the car.

Substituting the values in the above equation,

\(\begin{array}{c}{K_{\rm{T}}} = 1100{\rm{ kg}} \times {\left( {23.61{\rm{ m/s}}} \right)2}\\ = 613175.31{\rm{ kg}} \cdot {{\rm{m}}2}{\rm{/}}{{\rm{s}}2}\left( {\frac{{1{\rm{ J}}}}{{1{\rm{ kg}} \cdot {{\rm{m}}2}{\rm{/}}{{\rm{s}}2}}}} \right)\\ = 613175.31{\rm{ J }}\end{array}\).

The change in the entropy of the universe due to the result of this collision can be expressed as

\(\begin{array}{c}\Delta S = \frac{Q}{T}\\ = \frac{{{K_{\rm{T}}}}}{T}\end{array}\).

Here, T is the temperature at which the heat is transferred to the surrounding.

Substituting the values in the above equation,

\(\begin{array}{c}\Delta S = \frac{{613175.31{\rm{ J}}}}{{293{\rm{ K}}}}\\ = 2092.748{\rm{ J/K}}\end{array}\).

Thus, the change in the entropy of the universe is \(2092.748{\rm{ J/K}}\).