Question

A 1200 kg car traveling at 60 mph quickly brakes to a halt. The kinetic energy of the car is converted to thermal energy of the disk brakes. The brake disks (one per wheel) are iron disks with a mass of 4.0 kg.What is the increase in temperature of the brakes?

Answer 1

Answer:

 ΔT = 59.9 ° C

Explanation:

For this exercise the brake energy is totally converted into heat

Let's calculate the vehicle energy

       K = ½ m v²

Let's reduce the units to the SI system

       v = 30 mph (1609.34 m / mile) (1h / 3600s) = 13.41 m / s

      Em = K = ½ 1200 13.41²

      K = 1.079 105 J

All this energy is transformed into heat

     Em = Q

The expression for heat is

     Q = m  $c_{e}$ ΔT

     ΔT = Q / m $c_{e}$

The specific heat of iron is $c_{e}$ = 450 J / Kg ºC

     ΔT = 1,079 105 / (4.0 450)

     ΔT = 59.9 ° C

Answer 2

Answer:

Increase in temperature of the brakes = 59.94 K

Explanation:

Kinetic Energy: This is the energy of a body in motion. The Unit of kinetic Energy is Joules (J).

It is expressed mathematically as

Ek = 1/2mv²............... Equation 1

Ek = kinetic energy of the car, m = mass of the car, v = velocity of the car.

Given: m = 1200 kg, v = 60 mph = (60/3600) m/s = 26.82 m/s

Substituting these values into equation 1

Ek = 1/2(1200)(26.82)²

Ek = 1/2(1200)(719.31)

Ek = 600(719.31)

Ek = 431586 J

The kinetic energy from the car is converted to thermal energy of the disk brakes

Q = Ek = cm₁ΔT................................. Equation 2

Making ΔT the subject of the equation,

ΔT = Q/cm........................ Equation 3

ΔT  = increase in temperature of the brakes, m = mass of the brakes, c = specific heat capacity of the iron brake, Q = thermal energy of the brake disk

Given: m₁ = 4.0 kg × 4 = 16 kg, Q = 431586 J

Constant: c = specific heat capacity of iron = 450 J/kg.K

Substituting these values into equation 3,

ΔT  =431586/(16×450)

ΔT  = 431586/(7200)

ΔT = 59.94 K

Increase in temperature of the brakes = 59.94 K