When a driver brakes an automobile, the friction between the brake drums and the brake shoes converts the car's kinetic energy to thermal energy. if a 1500-kg automobile traveling at 35 m/s comes to a halt, how much does the temperature rise in each of the four 8.0 kg iron brake drums? (the specific heat of iron is 448 j/(kg · °

64.1°C

Explanation:

mass of automobile, M = 1500 kg

velocity, v = 35 m/s

mass of each brake, m = 8 kg

specific heat of iron, c = 448 J/kg°C

According to the conservation of energy

the kinetic energy of automobile = thermal energy of brakes

0.5 Mv² = 4 x m x c x ΔT

where, ΔT is the rise in temperature of the brake

0.5 x 1500 x 35 x 35 = 4 x 8 x 448 x ΔT

ΔT = 64.1°C

Thus, the change in temperature of each brake is 64.1°C.

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