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 1450-kg automobile traveling at 36 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).)

Respuesta :

Answer:

ΔT = 66 °C

Explanation:

The kinetic energy (Ek) of the car can be calculated using the following expression.

Ek = 1/2 × m × v² = 1/2 × (1450 kg) × (36 m/s)² = 9.4 × 10⁵ J

where,

m: mass

v: speed

When the car stops, this energy is converted into heat (Q) used to raise the temperature of the iron brake drums, according to the following expression.

Q = c × m × ΔT

where,

c: specific heat

m: mass

ΔT: change in the temperature

Q = c × m × ΔT

9.4 × 10⁵ J = (448 J/kg.°C) × (4 × 8.0 kg) × ΔT

ΔT = 66 °C

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