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#