Respuesta :
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 [tex]c_{e}[/tex] ΔT
ΔT = Q / m [tex]c_{e}[/tex]
The specific heat of iron is [tex]c_{e}[/tex] = 450 J / Kg ºC
ΔT = 1,079 105 / (4.0 450)
ΔT = 59.9 ° C
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
