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 1220 kg automobile traveling at 49.8 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
When the brakes of the automobile is applied it comes to halt. During this process the mechanical energy of the automobile is converted to heat energy in the brake pads. The automobile weighs 1220 kg and is initially moving at 49.8 m/s. It has 4 disk brakes that are applied together to bring it to a halt.
The mass of the steel brake disks is 3.5*4 = 14 kg. The heat capacity of steel is 488 J/kg*C which means that 488 J of heat increases the temperature of 1 kg of steel by 1 degree C.
The mechanical energy of the automobile was initially (1/2)*1220*(49.8)^2 = 1512824 J
All the energy is converted to heat in the disk brakes. This raises their temperature by 1512824/(488*14) = 221.4 C
The temperature of the disks in the brakes rises by 221 C to bring the automobile to a halt.