Question

A 750-kg car moving at 23 m/s brakes to a stop. Assume that all the kinetic energy is transformed into thermal energy. The brakes contain about 15 kg of iron, which absorbs the energy. What is the increase in temperature of the brakes?

Answer 1

Answer:

The temperature of the brakes is 29.38°C.

Explanation:

Given that,

Mass of car = 750 kg

Speed = 23 m/s

Mass of iron = 15 kg

We need to calculate the kinetic energy of car

Using formula of kinetic energy

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

$K.E=\dfrac{1}{2}\times750\times23^2$

$K.E=198.375\ kJ$

We need to calculate the temperature of the brakes

Using formula of energy

$K.E=mc\Delta T$

Put the value into the formula

$198.375=15\times0.450\times\Delta T$

$\Delta T=\dfrac{198.375}{15\times0.450}$

$\Delta T=29.38^{\circ}C$

Hence, The temperature of the brakes is 29.38°C.