Question

When a driver brakes an automobile, friction between the brake disks and the brake pads converts part of the car's translational kinetic energy to internal energy. If a 1260 kg automobile traveling at 30.8 m/s comes to a halt after its brakes are applied, how much can the temperature rise in each of the four 3.5 kg steel brake disks? Assume the disks are made of iron (cp = 448 J/kg·°C) and that all of the kinetic energy is distributed in equal parts to the internal energy of the brakes.

°C rise in temperature

Answer 1

Mass of the automobile = M = 1260 kg

Speed of the automobile = V = 30.8 m/s

Number of steel brake disk = n = 4

Mass of each steel brake disk = m = 3.5 kg

Specific heat of iron = C = 448 J/(kg.^oC)

Rise in temperature in each of the brake disk = T

The kinetic energy of the automobile is converted into heat energy of the brake disk as the automobile comes to a halt.

MV²/2 = nmCT

(1260)(30.8)²/2 = (4)(3.5)(448)T

T = 95.3 ^oC

Rise in temperature in each of the brake disk = 95.3 ^oC