Chapter 12, Problem 73A

To determine

Change in temperature of the brakes.

Answer to Problem 73A

  $\Delta T=11.61°\text{C}$

Explanation of Solution

Given:

Mass of automobile, $m_{car}=1500\text{kg}$

Final velocity of automobile, $v_f=0\text{m/s}$

Initial velocity of automobile, $v_i=25\text{m/s}$

Mass of aluminum brakes, $m_b=45\text{kg}$

Formula Used:

Change in kinetic energy is given as,

  $\Delta K=\frac{1}{2}m\left(v^2-v_o{}^2\right)$

Change in thermal energy is given as,

  $\Delta U=m_{b}C\Delta T$

From the law of thermodynamics,

  $Q_1+Q_2=0$

Where $Q_1$ and $Q_2$ is heat energy to different systems.

Calculation:

Here, $Q_1=\Delta K$ that is change in kinetic energy of car and $Q_2=\Delta U_T$ is the change in heat energy of aluminum brake.

As specific heat of aluminum is $C=897\text{J/kg}\text{.K}$

So,

  $\begin{array}{l}\Delta K+\Delta U_T=0\\ \frac{1}{2}{m}_{car}\left(v^2-v_o{}^2\right)+m_{b}C\Delta T=0\\ \Delta T=-\frac{\frac{1}{2}{m}_{car}\left(v^2-v_o{}^2\right)}{m_{b}C}\\ \Delta T=-\frac{\frac{1}{2}\times 1500\times \left(0^2-{25}^2\right)}{45\times 897}\\ \Delta T=11.61°\text{C}\end{array}$

Conclusion:

Therefore, the change in temperature of the brake is $\Delta T=11.61°\text{C}$ .