Question

The temperature of air changes from 0 to 18°C while its velocity changes from zero to a final velocity, and its elevation changes from zero to a final elevation. At which values of final air velocity and final elevation will the internal, kinetic, and potential energy changes be equal? The constant-volume specific heat of air at room temperature is 0.718 kJ/kg·ºC

Answer 1

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:

• Internal Energy: 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 $U=mC_{V}\Delta T$ where $m$ is the mass (kg), $C_{V}$ is a specific constant-volume (kJ/kg°C) and $\Delta T$ is the Temperature change in °C.

• Kinetic Energy: denotes the work done on an object (of given mass $m$) so that the object at rest, can accelerate to reach a final velocity. In S.I. is defined as $K=\frac{1}{2}mv^2$ where $v$ is the velocity of the object in (m/s).
• Potential Energy: denotes the energy occupied by an object (of given mass $m$) due to its position with respect to another object. In S.I. is defined as $P=mgh$, where $g$ is the gravity constant equal to $9,81m/s^2$ and $h$ is the elevation (meters).

Note: The Internal energy is unaffected by the Kinetic and Potential Energies.

Given Information:

• Temperature Change 0°C → 18°C ( thus $\Delta T=18$°C )
• Object velocity we shall call it $v_{o}$ and $v_{f}$, for initial and final, respectively. Here we also know that $v_{o}=0m/s^2$
• Object elevation we shall call it $h_{o}$ and $h_{f}$, for initial and final, respectively. Here we also know that $h_{o}= 0m$

∴ We are trying to find $v_{f}$ and $h_{f}$ of the air where $U$, $K$ and $P$ are equal.

Lets look at the change in Energy for each.

Step 1: Change in Kinetic Energy=Change in Internal Energy

$\Delta E_{K}=\Delta U\\\frac{1}{2}m{v_{f}}^2- \frac{1}{2}m{v_{o}}^2=mC_{V}\Delta T$

Here we recall that $v_{o}=0m/s^2$ and mass $m$ is the same everywhere. Thus we have:

$\frac{1}{2}m{v_{f}}^2=mC_{V}\Delta T$    

$\frac{1}{2} {v_{f}}^2=C_{V}\Delta T\\ {v_{f}}^2=2C_{V}\Delta T\\ v_{f}=\sqrt{2C_{V}\Delta T}$     Eqn(1)

Step 2: Change in Potential Energy=Change in Internal Energy

$\Delta E_{P}=\Delta U\\mgh_{f}-mgh_{o}=mC_{V}\Delta T$

Here we recall that $h_{o}=0m/s^2$ and mass $m$ is the same everywhere. Thus we have:

$mg(h_{f}-h_{o})=mC_{V}\Delta T\\gh_{f}=C_{V}\Delta T$

$h_{f}=\frac{C_{V}\Delta T}{g}$      Eqn(2).

Finally by plugging the known values in Eqns (1) and (2) we obtain:

$v_{f}=\sqrt{2*718*18}=160.77m/s$

$h_{f}=\frac{718*18}{9.81}=1,317.43m$

Thus we can conclude that for the air final velocity $v_{f}=160.77m/s$ and final elevation $h_{f}=1,317.43m$ the internal, kinetic, and potential energy changes  will be equal.