Question

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)

Answer 1

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`