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 1750-kg automobile traveling at 28 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 · °C.)

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Mv²/2 =4mcΔT
ΔT = Mv²/8mc=1750•28²/8•8•448=47.9℃

answered by Elena

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Question

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 1750-kg automobile traveling at 28 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 · °C.)

1 answer

Mv²/2 =4mcΔT
ΔT = Mv²/8mc=1750•28²/8•8•448=47.9℃

Ask a New Question or answer this question.

Elena · Answer

Mv²/2 =4mcΔT ΔT = Mv²/8mc=1750•28²/8•8•448=47.9℃