# A 5 g coin is dropped from a 300 m building. If it reaches a terminal velocity of 45 m/s, and the...

## Question:

A 5 g coin is dropped from a 300 m building. If it reaches a terminal velocity of 45 m/s, and the rest of the energy is converted to heating the coin, what is the change in temperature (in celsius) of the coin? (The specific heat of copper is 387 J/kg C)

## Conservation of Energy:

When we have a closed system, we consider no energy to be able to flow in or out of it. This means that whatever energy was given to the system, it would be maintained. The total energy would now be constant and any motion inside the system would change its kinetic energy and thus the potential energy must adjust accordingly to maintain a constant total energy.

Given:

• {eq}\displaystyle m = 5\ g = 0.005\ kg {/eq} is the mass of the coin
• {eq}\displaystyle h = 300\ m {/eq} is the height of the building
• {eq}\displaystyle v = 45\ m/s {/eq} is the terminal speed of the coin
• {eq}\displaystyle c = 387\ J/kg^\circ C {/eq} is the specific heat of copper

Before the coin is dropped, all of its energy is equal to a gravitational potential energy. Now any energy change must come from this constant total energy and at the bottom of the building, we expect that all of the energy is split between the kinetic energy due to the terminal speed and the heat gained by the coin. We can thus write our energy conservation equation as:

{eq}\displaystyle mgh = \frac{1}{2} mv^2 + mc\Delta T {/eq}

We cancel the mass terms;

{eq}\displaystyle gh = \frac{1}{2} v^2 + c\Delta T {/eq}

We isolate the change in temperature as:

{eq}\displaystyle c\Delta T = gh - \frac{v^2}{2} {/eq}

{eq}\displaystyle \Delta T = \frac{2gh - v^2}{2c} {/eq}

We substitute:

{eq}\displaystyle \Delta T = \frac{2(9.8\ m/s^2)(300\ m) - (45\ m/s)^2}{2(387\ J/kg^\circ C)} {/eq}

We simplify this expression and we will thus get:

{eq}\displaystyle \boxed{\Delta T = 4.98^\circ C} {/eq}