   Chapter 11, Problem 24P

Chapter
Section
Textbook Problem

When a driver brakes an automobile, the friction between the brake drams and the brake shoes converts the car’s kinetic energy to thermal energy. If a 1 500-kg automobile traveling at 30 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 448J/kg · °C.)

To determine
The how much does the temperature rise in each of the four iron brake drum.

Explanation

Given Info: Mass of the automobile is 1500 kg, initial speed of automobile 30 m/s, final speed of automobile 0, mass of the iron brake drum is 8.0kg, specific heat of iron is 448J/kg°C .

The change in kinetic energy of the car as the car comes to halt from motion is transferred to the 4 brakes of the car and the internal energy of the brakes increases.

Formula to calculate the change in kinetic energy of the car is,

ΔKE=12m(vf2vi2)

ΔKE is the change in kinetic energy

• m is the mass of the car,
• vi is the initial speed of the car,
• vf is the final speed of the car,

Formula to calculate Heat gained by break drum is,

QFe=4mFecFe(ΔT)

• QFe is the heat gained by break drum,
• mFe is the mass of a iron break drum,
• ΔT is the increase in temperature of iron break drum,

4 represents 4 breaks drums present in the car.

Change in kinetic energy of the car is equal to the increase in internal energy of the 4 brakes.

ΔKE=QFe .

Equate both energy equation and to rewrite in terms of ΔT

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