Chapter 12, Problem 59A

To determine

The temperature increase in the iron brakes.

Answer to Problem 59A

The temperature increase in the iron brakes is $290.86°\text{C}$.

Explanation of Solution

Given:

A car of mass 750-kg moving with a velocity of 23 m/s applies brakes to stop and the kinetic energy of the car is converted into thermal energy exclusively absorbed by the brakes which is made of iron and has a mass of 15 kg. This thermal energy increases the temperature of the brakes.

Formula Used:

According to First Law of thermodynamics, the thermal energy change of an object is,

  $\Delta U=Q-W$

Where,

Q is the heat supplied to the object

W is the work done by the object.

Now, when the system is isolated from the surroundings, the thermal energy is

  $\Delta U=-W$........(1)

Here, the work done by the car is converted to the thermal energy absorbed by the brakes.

The work done by the car of mass m is its change in kinetic energy, expressed as,

  $\begin{array}{c}{W}_{car}=\Delta K.E\\ =\frac{1}{2}{m}_{car}{\left(\Delta v\right)}^2\end{array}$

In the above equation, the parameter $\Delta v$ represents the change in velocity of the car.

The energy for causing temperature change in the brakes is dependent on the temperature change ,mass of the brakes $m_{brakes}$ and specific heat $C_{iron}$ , and is given by the expression,

  $\Delta U_{brakes}=m_{brakes}{C}_{iron}\Delta T$

Calculation:

Here, car is having a mass of $m\text{=750- kg}$ .The car moves with a velocity $v\text{=23 m/s}$ and comes to rest making its final velocity to be zero. Now, the change in kinetic energy of the car is given by,

  $\begin{array}{c}\Delta K.E=\frac{1}{2}m{\left(\Delta v\right)}^2\\ =\frac{1}{2}\times 750\times {23}^2\\ =19.8375\times {10}^4\text{ J}\end{array}$

Now, thermal energy of the iron brakes with specific heat $C_{iron}=450\text{ J/kg}\text{.K}$ is

  $\begin{array}{c}\Delta U_{brakes}=m_{brakes}{C}_{iron}\Delta T\\ =15\times 450\times \Delta T\\ =6750\Delta T\end{array}$

Now, the change in Kinetic energy is converted into thermal energy and hence it follows that,

  $\begin{array}{c}\Delta U=\Delta K.E\\ 6750\Delta T=19.8375\times {10}^4\text{ J}\\ \Delta T=\frac{19.8375\times {10}^4\text{ J}}{6750}\\ =29.38°\text{C}\end{array}$

Conclusion:

Therefore, the temperature increase in the brakes is $29.38°\text{C}$ .