When a driver brakes an automobile, friction between the brake disks and the bra

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 1274 kg automobile traveling at 28 m/s comes to a halt after its brakes are applied, how much can the temperature rise in each of the four 3.1 kg steel brake disks? Assume the disks are made of iron (with specific heat 448 J/kg- degree C) and that all of the kinetic energy is distributed in equal parts to the internal energy of the brakes. Answer in units of degree C

Explanation / Answer

K.E. = m*CI*T

=>T = (0.5*1274*28*28)/(4*3.1*448) = 89.9 0C

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