A 2000 kg car is traveling at 20.0 m/s. Then, it puts on the brakes and comes to a complete stop. The heat generated due to the stopping heats up the brakes. The mass of the brake system is 20.0 kg and it has a specific heat of 0.200 kcal/kg °C. What is the increase in the temperature of the brakes? ( 1 cal = 4.186 J)
kinetic energy = (1/2) m v^2
that goes into heating the brakes
(use Joules not calories)
energy
=specific heat*mass*(Tend-Tbegin)
To calculate the increase in temperature of the brakes, we can use the formula:
Q = mcΔT
Where:
Q is the heat generated
m is the mass of the brake system
c is the specific heat of the brakes
ΔT is the increase in temperature
First, let's calculate the heat generated (Q):
Q = KE
Where:
KE is the kinetic energy of the car
The formula for kinetic energy is:
KE = 0.5mv^2
Where:
m is the mass of the car
v is the velocity of the car
Plugging in the given values:
KE = 0.5 * (2000 kg) * (20.0 m/s)^2
KE = 0.5 * 2000 kg * 400 m^2/s^2
KE = 400000 J
Now, let's calculate the increase in temperature (ΔT) using the formula:
Q = mcΔT
ΔT = Q / (mc)
Plugging in the given values:
ΔT = 400000 J / (20.0 kg * 0.200 kcal/kg °C * 4.186 J/cal)
ΔT = 400000 J / (8.372 J/°C)
ΔT ≈ 47.77 °C
Therefore, the increase in temperature of the brakes is approximately 47.77 °C.
To find the increase in temperature of the brakes, we need to calculate the heat generated during the deceleration of the car and then use the specific heat capacity of the brakes to determine the temperature change.
First, let's calculate the heat generated during the deceleration of the car using the principle of conservation of energy. The initial kinetic energy (KE1) of the car can be calculated using the formula:
KE1 = (1/2) * mass * velocity^2
where mass is the mass of the car (2000 kg) and velocity is the initial velocity (20.0 m/s).
KE1 = (1/2) * 2000 kg * (20.0 m/s)^2
= 400,000 J
Since the car comes to a complete stop, all of the initial kinetic energy is converted into heat. So, the heat generated (Q) is equal to the initial kinetic energy:
Q = 400,000 J
Now, we can use the specific heat capacity (c) of the brake system, the mass of the brake system (20.0 kg), and the equation:
Q = m * c * ΔT
where ΔT is the change in temperature.
To convert the specific heat capacity from kcal/kg °C to J/kg °C, we need to use the conversion factor. Since 1 cal = 4.186 J, we have:
c = 0.200 kcal/kg °C * 4.186 J/cal
= 0.8372 J/kg °C
Substituting the values into the equation, we can solve for ΔT:
400,000 J = 20.0 kg * 0.8372 J/kg °C * ΔT
ΔT = 400,000 J / (20.0 kg * 0.8372 J/kg °C)
= 23.96 °C
Therefore, the increase in the temperature of the brakes is approximately 23.96 °C.