How Does Braking Affect the Temperature of Car Brakes?

[Cortar]

A 758 kg car moving at 25 m/s brakes to a
stop. The brakes contain about 16 kg of iron
that absorb the energy.
What is increase in temperature of the
brakes? Assume the specific heat of iron is
450 J=kg ¢
± C. Answer in units of ±C.Q=mc(t1-t2)

Ok I have the mass of the brakes and the specific heat and I know you only have to solve for the change in temperature, so I all need is Q.
I know Q is measured in Joules and J=Newton x distance, but all I am given is the velocity of the car, nothing about time, distance, or Acceleration. So how do you figure the Q?

 

[kdv]

A 758 kg car moving at 25 m/s brakes to a
stop. The brakes contain about 16 kg of iron
that absorb the energy.
What is increase in temperature of the
brakes? Assume the specific heat of iron is
450 J=kg ¢
± C. Answer in units of ±C.


Q=mc(t1-t2)

Ok I have the mass of the brakes and the specific heat and I know you only have to solve for the change in temperature, so I all need is Q.
I know Q is measured in Joules and J=Newton x distance, but all I am given is the velocity of the car, nothing about time, distance, or Acceleration. So how do you figure the Q?

It comes to a stop so all the initial kinetic energy (1/2 mv^2) goes into heat. That's your Q

 

[Moetasim]

To calculate the energy absorbed by the brakes (Q), we can use the formula for kinetic energy: KE = 1/2mv^2. In this case, the initial kinetic energy of the car is converted into thermal energy in the brakes. So, the energy absorbed by the brakes can be calculated as:

Q = KE = 1/2mv^2 = 1/2(758 kg)(25 m/s)^2 = 237,250 J

Now, we can use the formula Q = mcΔT to calculate the change in temperature (ΔT) of the brakes. Rearranging the formula, we get:

ΔT = Q/mc = (237,250 J)/(16 kg)(450 J/kg °C) = 33.15 °C

Therefore, the increase in temperature of the brakes is approximately 33.15 °C.