Answer:
For the air:
Final Velocity 160.77m/s
Final Elevation 1,317.43m
the Internal, Kinetic, and Potential Energy changes will be equal.
Explanation:
In principle we know the following:
- <u>Internal Energy:</u> is defined as the energy contained within a system (in terms of thermodynamics). It only accounts for any energy changes due to the internal system (thus any outside forces/changes are not accounted for). In S.I. is defined as where is the mass (kg), is a specific constant-volume (kJ/kg°C) and is the Temperature change in °C.
- <u>Kinetic Energy:</u> denotes the work done on an object (of given mass ) so that the object at rest, can accelerate to reach a final velocity. In S.I. is defined as where is the velocity of the object in (m/s).
- <u>Potential Energy:</u> denotes the energy occupied by an object (of given mass ) due to its position with respect to another object. In S.I. is defined as , where is the gravity constant equal to and is the elevation (meters).
<em>Note: The Internal energy is unaffected by the Kinetic and Potential Energies.</em>
<u>Given Information:</u>
- Temperature Change 0°C → 18°C ( thus °C )
- Object velocity we shall call it and , for initial and final, respectively. Here we also know that
- Object elevation we shall call it and , for initial and final, respectively. Here we also know that
∴<em> We are trying to find and of the air where , and are equal.</em>
Lets look at the change in Energy for each.
<u>Step 1: Change in Kinetic Energy=Change in Internal Energy</u>
Here we recall that and mass is the same everywhere. Thus we have:
Eqn(1)
<u>Step 2: Change in Potential Energy=Change in Internal Energy</u>
Here we recall that and mass is the same everywhere. Thus we have:
Eqn(2).
Finally by plugging the known values in Eqns (1) and (2) we obtain:
Thus we can conclude that for the air final velocity and final elevation the internal, kinetic, and potential energy changes will be equal.