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 · °

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