# Insulated box of air at temperature $T$, is dropped from height $H$, what is the temperature of the air inside box on the ground? [closed]

In an effort to understand more about the first law I was wondering about the potential energy term and how it would influence the gas inside the box I made this question and tried to answer it.

I will assume that the air follows the ideal gas law, there is no air resistance and the box stops moving immediately when it reaches the rigid ground.

The first law: $$0 = \Delta PE + \Delta U\\ gH = c_v (T' - T)\\ T' = gH/c_v + T$$

would this be correct or missing something? (i.e. invalid assumptions..)

I am also wondering if the energy is fully transferred to the ground instead..

• Gravitational potential energy is converted into kinetic energy, not heat.
– user122423
Jun 14 '17 at 2:59
• Jun 14 '17 at 6:33

$Energy\space of\space the\space system\space (E) = Internal\space energy\space (U) + PE_{system} + KE_{system}$,

$\Delta E =\Delta U+\Delta{PE} + \Delta{KE}$,

The condition for $\Delta E =\Delta{U}$ is $\Delta{PE}=\Delta{KE}=0$.

The first law of thermodynamics gives us the relation between the change in energy of the system and heat and work.

$\Delta E + PdV=Q$,

$\Delta U+\Delta {PE} + PdV=Q$ $\tag1$

In your question, $\Delta {PE}=-mgh$ and $PdV=0$ $(as\space dV=0)$.

Exactly, $dU=\delta Q -\delta W$, and here $\delta W=mgh$ and $\delta Q=0$. You can run some numbers to see that the value is very small for realistic settings.