When a driver brakes an automobile, friction between the brake disks and the brake pads converts part of the car's translational kinetic energy to internal energy. If a 1250 kg automobile traveling at 37.2 m/s comes to a halt after its brakes are applied, how much can the temperature rise in each of the four 3.5 kg steel brake disks? Assume the disks are made of iron (cp = 448 J/kg·°C) and that all of the kinetic energy is distributed in equal parts to the internal energy of the brakes.
To find out how much the temperature can rise in each of the brake disks, we need to calculate the change in internal energy.
The change in internal energy can be calculated using the equation:
ΔE = mcΔT
where ΔE is the change in internal energy, m is the mass of the object, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, we have four brake disks, each with a mass of 3.5 kg. The specific heat capacity of iron is given as cp = 448 J/kg·°C.
First, we need to calculate the initial kinetic energy of the car:
KE = (1/2)mv^2
where m is the mass of the car and v is its velocity.
Plugging in the values, we have:
KE = (1/2)(1250 kg)(37.2 m/s)^2
Next, since all of the kinetic energy is distributed equally among the four brake disks, we need to divide the initial kinetic energy by four:
KE_disk = (1/4) KE
Now, we can calculate the change in internal energy for each brake disk using the equation mentioned above:
ΔE_disk = mcΔT
Plugging in the values, we have:
ΔE_disk = (3.5 kg)(448 J/kg·°C)ΔT
Since the change in internal energy is equal to the initial kinetic energy of each disk (ΔE_disk = KE_disk), we can equate the two equations:
(3.5 kg)(448 J/kg·°C)ΔT = (1/4) KE
Now, we can solve for ΔT:
ΔT = ( (1/4) KE ) / (3.5 kg)(448 J/kg·°C)
Substituting the values for KE and solving:
ΔT = ( (1/4) (1250 kg)(37.2 m/s)^2 ) / (3.5 kg)(448 J/kg·°C)
Calculating this expression will give us the change in temperature (ΔT) that each brake disk will experience.