Physics, 29.10.2019 21:31 lizzie3545

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

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Answer from: james169196

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

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

) 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

is the velocity of the object in (m/s). Potential Energy: 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 P=mgh

, where

is the gravity constant equal to 9,81m/s^2

and

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 , and 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 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 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.

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