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

For the air:

Final Velocity 160.77m/s

Final Elevation 1,317.43m

the Internal, Kinetic, and Potential Energy changes will be equal.

Step-by-step 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=(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\\(1)/(2)m{v_(f)}^2- (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:

`(1)/(2)m{v_(f)}^2=mC_(V)\Delta T`

`(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)=(C_(V)\Delta T)/(g)` Eqn(2).

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

`v_(f)=√(2*718*18)=160.77m/s`

`h_(f)=(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.