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.

Answer from: Quest

a high pressure system is a dense air mass that is usually cooler and drier than the surrounding air. a low pressure system is a less dense air mass that is usually wetter and warmer than the surrounding air.

in general, areas that experience high atmospheric pressure also experience fair weather. low pressure systems can cause the formation of clouds and storms. air usually flows from areas of high pressure to areas of low pressure..

Answer from: Quest

im srry but ur wrong the answer is d a smooth rock, some red dirt, and an empty bird's nest.

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The temperature of air changes from 0 to 18°c while its velocity changes from zero to a final veloci...