Question

A 763 kg car moving at 26 m/s brakes to a stop. The brakes contain about 15 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.

Answer 1

`\Delta T=38.20^(\circ)`

Step-by-step explanation:

It is given that,

Mass of the car, m = 763 kg

Speed of the car, v = 26 m/s

Mass of the iron, m' = 15 kg

Specific heat of iron, c = 450 J/kg

When the car is in motion, it will possess kinetic energy. It is given by :

`K=(1)/(2)mv^2`

`K=(1)/(2)* 763* (26)^2`

K = 257894 J

Since, energy is absorbed by the brakes. The kinetic energy of the car is absorbed by the brakes. So,

`K=mc\Delta T`

`\Delta T` is the increase in temperature of the brakes

`\Delta T=(K)/(m'c)`

`\Delta T=(257894)/(15* 450)`

`\Delta T=38.20^(\circ)`

So, the increase in temperature of the brakes is 38.20 degrees Celsius. Hence, this is the required solution.