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If we take the first law of thermodynamics: $$ΔQ = ΔU+ΔW$$ And we consider a system of a ball falling from height $h$ in an Earth-like gravitational field(no air drag and $h$<<$Rₑ$) $$ΔU = mgh$$ and $$ΔW = -mgh$$ therefore ΔQ = 0 and the process must be adiabatic.

I think I am going wrong somewhere as this means that the ball on falling and just before hitting the ground has a higher temperature than when it was released (even if atmospheric temperature is same/ nearly same at both points). How is that possible?

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  • $\begingroup$ @JohnDoty the question framing removes the effect of friction/drag of any kind. $\endgroup$
    – TheTheoMess
    Commented Feb 23 at 16:22
  • $\begingroup$ "heating up" and "warming up" are not the same: "warming up" is the increase of temperature, "heating up" is the absorption of thermal work, these can be related or unrelated just as a turbine can be turned by electric work from a battery or by thermal effects. $\endgroup$
    – hyportnex
    Commented Feb 23 at 16:33

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You are making many mistakes.

Before a proto-star gets hot enough to start the nuclear fusion reaction to produce light, it still needs to heat up. Gravitational attraction is the driving force for this, for it is not sufficient alone to heat things up. You need electrodynamics, i.e. a 2nd interaction, to convert the energy liberated by going deeper into a gravitational well, into heat, so that the conditions of nuclear fusion can be reached.

As for your particular paradox, it is just that you are using thermodynamics equations that assume that the centre of mass is stationary. If you worked out to higher detail and included the kinetic energy of the centre of mass, then the ball would have transferred the energy from gravitational potential energy to ordered kinetic energy of the centre of mass, so that the ball is cold just before hitting the ground. Upon hitting the ground, some of that ordered kinetic energy of the centre of mass can be converted to heat that raises the temperature.

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In the first law equation for a closed system

$$\Delta U=Q-W$$

$\Delta U$ is the change in internal energy of the system. That is the change in molecular kinetic and potential energy of the system with respect to its center of mass. It does not include changes in gravitational potential energy or kinetic energy of the center of mass of the system with respect to an external (to the system) frame of reference. For that, you need to use the more general form of the first law, i.e.,

$$Q-W=\Delta U+\Delta KE+\Delta PE$$

The last two terms apply to the motion and position of the system as a whole, in this case, the system of the ball plus Earth.

Hope this helps.

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As @BobD points out, the general form of the first law of thermodynamics reads:$$\Delta U+\Delta KE+\Delta PE=Q-W$$In the case of a falling ball before it hits, there is no heat flow to the ball, and no deformational work is done by the ball, and its temperature doesn't change, so we are left with $$\Delta KE+\Delta PE=0$$In the case where the ball hits the ground, is no longer moving, and comes to thermal equilibrium with the surroundings, we have $$\Delta PE=Q$$Virtually all the potential energy of the ball is changes to heat transferred tot he surroundings.

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