Question

The brakes in a car increase in temperature by\({\bf{\Delta T}}\)when bringing the car to rest from a speed\(v\). How much greater would\({\bf{\Delta T}}\)be if the car initially had twice the speed? You may assume the car to stop sufficiently fast so that no heat transfers out of the brakes.

Short answer

The value of \({\rm{\Delta T}}\) will be four times greater than the initial value if the speed is twice the initial value.

Step-by-step solution

Describe the scenario

Assume that there is no heat transfer occurring in this situation. The amount of heat transfer can be calculated using the expression;

\({\rm{Q = mc\Delta T}}\)

Here, \({\rm{Q}}\)is the amount of heat, \({\rm{m}}\)is the mass of the substance, \({\rm{c}}\)is the specific heat capacity, and\({\rm{\Delta T}}\)is the variation in temperature.

Here, the value \({\rm{Q}}\)is zero because there is no heat transfer. For this, \({\rm{\Delta T}}\)also have to be zero as well. That is, the temperature remains constant. In this situation, the speed decreased. In order to maintain the temperature at the rest position, breaks in the car increase the temperature.

Describe the relation between temperature and speed

The temperature is the measure of theaverage kinetic energy of atoms in a material. Kinetic energy is proportional to the square of velocity. So, the temperature will also be proportional to the square of the velocity. If the speed doubles, the temperature will become four times higher than the initial temperature.

So, \({\rm{\Delta T}}\) will be four times as large if the speed becomes twice the initial speed.