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.

Question

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.

Answer 1

KE=heatgained

1/2 1250 37.2^2=3.5*ciron*deltatTemp
solvefor delta temp

Answer 2

To find the temperature rise in each of the four steel brake disks, we can use the following steps:

Step 1: Calculate the initial kinetic energy of the car.
The kinetic energy (KE) can be calculated using the formula:
KE = 0.5 * mass * velocity^2

Given:
mass of the car (m) = 1250 kg
velocity (v) = 37.2 m/s

Using the formula,
KE = 0.5 * 1250 kg * (37.2 m/s)^2

Step 2: Calculate the total kinetic energy converted to internal energy.
Since all of the kinetic energy is distributed in equal parts to the internal energy of the brakes, we need to divide the initial kinetic energy by the number of brake disks (4 in this case) to find the energy for each brake disk.

Total energy for each brake disk = KE / 4

Step 3: Calculate the temperature rise in each brake disk.
The equation for heat energy (Q) is given by:
Q = mass * specific heat capacity * temperature change

In this case, the heat energy (Q) is equal to the energy converted to internal energy for each brake disk.

Given:
mass of each brake disk (m_brake) = 3.5 kg
specific heat capacity of iron (cp) = 448 J/kg·°C

Using the equation:
Q = m_brake * cp * ΔT

Rearranging the equation to solve for temperature change (ΔT):
ΔT = Q / (m_brake * cp)

Step 4: Substitute the values and calculate the temperature rise.
Substituting the values into the equation:
ΔT = (Total energy for each brake disk) / (m_brake * cp)

Finally, calculate the temperature rise in each brake disk by substituting the values and making the calculation.

Answer 3

To determine the temperature rise in each brake disk, we can use the equation:

ΔE = mcΔT

Where:
ΔE is the change in internal energy (in joules),
m is the mass of the object (in kilograms),
c is the specific heat capacity of the material (in joules per kilogram per degree Celsius), and
ΔT is the change in temperature (in degrees Celsius).

Let's calculate the change in internal energy first:

ΔE = (1/2)mv^2

Where:
m is the mass of the car (in kilograms),
v is the velocity of the car (in meters per second).

Substituting the given values:

ΔE = (1/2)(1250 kg)(37.2 m/s)^2

ΔE ≈ 869700 J

Since the kinetic energy is distributed equally to all four brake disks, we divide ΔE by 4 to find the energy transferred to each disk:

ΔE_disk = ΔE / 4

ΔE_disk ≈ 217425 J

Now we can calculate the temperature rise in each brake disk using the specific heat capacity (cp) of iron:

ΔE_disk = m_disk * cp * ΔT_disk

Rearranging the equation to solve for ΔT_disk:

ΔT_disk = ΔE_disk / (m_disk * cp)

Substituting the given values:

ΔT_disk = 217425 J / (3.5 kg * 448 J/kg·°C)

ΔT_disk ≈ 175 °C

Therefore, the temperature can rise approximately 175 degrees Celsius in each of the four 3.5 kg steel brake disks.