A 750kg car moving at 23m/s brakes to a stop. The brakes contain about 15.0kg of iron that absorbs the energy. What is the increase in temperature of the brakes? (HINT: KEcar=Qbrakes) (ANSWER: 30 degrees Celsius)

1 Answer
Feb 1, 2018

The increase in temperature is =29.4^@C

Explanation:

The kinetic energy of the car is KE=1/2mv^2

The mass of the car is m=750kg

The velocity of the car is v=23ms^-1

The kinetic energy is

KE=1/2*750*23^2=198375J=198.375kJ

This energy is absorbed by the brakes.

The specific heat of iron is C_(Fe)=0.45kJkg^-1K^-1

The mass of iron is m_(Fe)=15kg

Let the difference in temperature be DeltaT

Therefore,

m_(Fe)*C_(Fe)*DeltaT=KE

15*0.45*DeltaT=198.375

DeltaT=198.375/(15*0.45)=29.4^@C