Question

In braking an automobile, the friction between the brake drums and brake shoes converts the car's kinetic energy into heat. If a 1 500-kg automobile traveling at 30 m/s brakes to a halt, how much does the temperature rise in each of the four 8.0-kg brake drums in °C? (The specific heat of each iron brake drum is 448 J/kg⋅°C).

Answer 1

To solve this problem it is necessary to apply the concepts related to energy conservation.

In this case the kinetic energy is given as

$KE = \frac{1}{2} mv^2$

Where,

m = mass

v= Velocity

In the case of heat lost energy (for all 4 wheels) we have to

$Q = mC_p \Delta T \rightarrow 4Q = 4mC_p \Delta T$

m = mass

$C_p =$ Specific Heat

$\Delta T$= Change at temperature

For conservation we have to

$KE = Q$

$\frac{1}{2} mv^2 = 4mC_p \Delta T$

$\Delta T = \frac{1}{2}\frac{mv^2}{4mC_p}$

$\Delta T = \frac{1}{2}\frac{(1500)(30)^2}{4(8)(448)}$

$\Delta T = 47.084\°C \approx 47\°C$

Therefore the temperature rises in each of the four brake drums around to 47°C