Question

Short answer

The increase in temperature of the ball is \(0.14^\circ {\rm{C}}\).

Step-by-step solution

Given data

The initial velocity of the ball is\({v_{\rm{i}}} = 22\;{\rm{m/s}}\).

The final velocity of the ball is\({v_{\rm{f}}} = 12\;{\rm{m/s}}\).

The specific heat of the rubber is \(c = 1200\;{\rm{J/kg}} \cdot ^\circ {\rm{C}}\).

Understanding loss of kinetic energy

In this problem, to find the increase in temperature, consider that the loss of kinetic energy is converted into heat, which changes the temperature of the ball.

Calculation of the change in temperature

The relation to calculate temperature is given by:

\(\begin{aligned}{c}KE = Q\\\frac{1}{2}m{\left( {{v_{\rm{i}}} - {v_{\rm{f}}}} \right)^2} = mc\Delta T\\\frac{1}{2}{\left( {{v_{\rm{i}}} - {v_{\rm{f}}}} \right)^2} = c\Delta T\end{aligned}\)

Here,\(\Delta T\)is the change in temperature.

On plugging the values in the above relation, you get:

\(\begin{aligned}{c}\frac{1}{2}\left( {{{\left( {22\;{\rm{m/s}}} \right)}^2} - {{\left( {12\;{\rm{m/s}}} \right)}^2}} \right) = \left( {1200\;{\rm{J/kg}} \cdot ^\circ {\rm{C}}} \right)\Delta T\\\Delta T = 0.14^\circ {\rm{C}}\end{aligned}\)

Thus, \(\Delta T = 0.14^\circ {\rm{C}}\) is the required change in temperature.