Question

In braking an automobile, the friction between the brake drums and brake shoes converts the car's kinetic energy into heat. If a 1 500-kg automobile traveling at 30 m/s brakes to a halt, how much does the temperature rise in each of the four 8.0-kg brake drums in °C? (The specific heat of each iron brake drum is 448 J/kg⋅°C).

Answer 1

To solve this problem it is necessary to apply the concepts related to energy conservation.

In this case the kinetic energy is given as

`KE = (1)/(2) mv^2`

Where,

m = mass

v= Velocity

In the case of heat lost energy (for all 4 wheels) we have to

`Q = mC_p \Delta T \rightarrow 4Q = 4mC_p \Delta T`

m = mass

`C_p =` Specific Heat

`\Delta T`= Change at temperature

For conservation we have to

`KE = Q`

`(1)/(2) mv^2 = 4mC_p \Delta T`

`\Delta T = (1)/(2)(mv^2)/(4mC_p)`

`\Delta T = (1)/(2)((1500)(30)^2)/(4(8)(448))`

`\Delta T = 47.084\°C \approx 47\°C`

Therefore the temperature rises in each of the four brake drums around to 47°C