Question

A 5.8-kg box having an initial speed of 4.0 m/s slides along a rough table and comes to rest. Estimate the total change in entropy of the universe. Assume all objects are at room temperature (293 K).

Short answer

The change in entropy is \(0.16\;{\rm{J/K}}\).

Step-by-step solution

Concepts

When the object comes to rest, the total kinetic energy is converted into heat energy.

The change in entropy is\(\Delta S = \frac{Q}{T}\).

Given data

The mass of the box is \(m = 5.8\;{\rm{kg}}\).

The initial speed of the box is \(v = 4.0\;{\rm{m/s}}\).

The room temperature is \(T = 293\;{\rm{K}}\).

Calculation

When the box comes to rest, the initial kinetic energy is converted into heat energy.

Now the heat energy is \(Q = \frac{1}{2}m{v2}\).

Therefore the change in entropy is,

\(\begin{array}{c}\Delta S = \frac{Q}{T}\\ = \frac{{\frac{1}{2}m{v2}}}{T}\\ = \frac{{\frac{1}{2} \times \left( {5.8\;{\rm{kg}}} \right) \times {{\left( {4.0\;{\rm{m/s}}} \right)}2}}}{{293\;{\rm{K}}}}\\ = 0.16\;{\rm{J/K}}\end{array}\)

Hence, the change in entropy is \(0.16\;{\rm{J/K}}\).