Oh, sorry, I mistakenly referenced this equation for pressure lapse as Claussius-Clapeyron. Sorry for inconvenience.

@JeopardyTempest That's the equation I have been taught. Also, $\Delta H = H - H_0$ (change of height). $R_u=\frac{R}{M(air)}=287.053 \frac{m^2}{s^2 K}$. Your $Z$ in your version of the equation might be $\frac{g}{R_u T}=\frac{9.8}{287.053\cdot 288.15 m}=1.184\cdot 10^{-4} m^{-1}$ because 15 °C is the average temperature and 9.8 m/s^2 is the average gravitational acceleration. Is this correct?

Interesting question, but I believe that the energy gets transformed into the another form so rapidly that it is nearly impossible to accomplish this.

Ok, but the data should have more westerly winds, I believe.

Where did you get that wind rose? I found this (world-weather.info/archive/romania/medias) wind rose and there is much more westerly winds.

@JeopardyTempest Maybe he wanted to give the approximate function of heat flux. If he wanted to do that, it would be $P(t)=-\cos{t}$ and thus the integral $T(t)=-\sin{t}+T_0$. With this function we can see the minimum temperature at $\frac{\pi}{2}$ (= 6.00 a.m.) Otherwise, I think he has just overcomplicated the situation without making it better.

Is solar power really maximum at $\frac{\pi}{2}$? (= 6:00 a.m.) Think about this again.

@EricDuminil Yes, in the August, the mean cooling rate is equal to the mean warming rate, so the temperature is at its maximum.