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

Answer:

$\Delta T=38.20^{\circ}$

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 :

$K=\dfrac{1}{2}mv^2$

$K=\dfrac{1}{2}\times 763\times (26)^2$

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

$\Delta T=\dfrac{K}{m'c}$

$\Delta T=\dfrac{257894}{15\times 450}$

$\Delta T=38.20^{\circ}$

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