In white dwarf stars, no nuclear reactions occur, and their temperature decreases due to heat radiation. Assuming that their surface is a perfect black body and that the temperature distribution inside the star is always homogeneous, determine the time dependence of the temperature! (Data for the van Maanen star: mass $m=10^{33} \mathrm{~g}$, radius $r=5500 \mathrm{~km}$, average specific heat $c=0.5 \mathrm{kcal} / \mathrm{g} \cdot \mathrm{K}$, initial temperature $T_{0}=16700 \mathrm{~K}$. (According to the Stefan-Boltzmann law, the power radiated per unit area of a perfect black body is $\sigma T^{4}$, where $T$ is the absolute temperature, $\left.\sigma=5.57 \cdot 10^{-12} \mathrm{~W} / \mathrm{K}^{4} \mathrm{~cm}^{2}.\right)$
If the specific heat of a star is $c$, then its energy changes by the value

$$
\Delta U=c \cdot m \cdot \Delta T
$$

for a temperature change of $\Delta T$. This energy change is equal to the energy radiated during the time $\Delta t$, which, according to the Stefan-Boltzmann law, can be written as

$$
\Delta U=-A \sigma T^{4} \cdot \Delta t ; \quad\left(A=4 \pi r^{2}\right)
$$

The negative sign indicates that the energy decreases over time.

The value of $\Delta U$ calculated in two ways is equal:

$$
c \cdot m \cdot \Delta T=-A \sigma T^{4} \Delta t
$$

from which

$$
\frac{\Delta T}{\Delta t}=-\frac{A \sigma}{c m} T^{4}
$$

Since $\lim _{\Delta t \rightarrow 0}(\Delta T / \Delta t)$ is the derivative of the temperature as a function of time ( $\mathrm{d} T / \mathrm{d} t$ ), we get from the above equation, taking the limit as $\Delta t \rightarrow 0$:

$$
\mathrm{d} T / \mathrm{d} t=-(A \sigma / c m) T^{4}
$$

From this differential equation, the desired function $T(t)$ can be obtained, taking into account the initial condition $T(0)=T_{0}$. The differential equation can be written as:

$$
\begin{gathered}
\frac{1}{T^{4}} \frac{\mathrm{d} T}{\mathrm{~d} t}=-\frac{A \sigma}{c m} \\
-\frac{1}{3} \frac{\mathrm{d}}{\mathrm{d} t}\left(T^{-3}\right)=-\frac{A \sigma}{c m}, \quad \frac{\mathrm{d}}{\mathrm{d} t}\left(T^{-3}\right)=\frac{3 A \sigma}{c m}
\end{gathered}
$$

thus

$$
\begin{gathered}
T^{-3}=\frac{3 A \sigma}{c m} t+C \\
T=\left(\frac{3 A \sigma}{c m} t+C\right)^{-1 / 3}
\end{gathered}
$$

From the condition $T(0)=T_{0}$, it follows that $C=T_{0}^{-3}$, so

$$
T=\left(\frac{3 A \sigma}{c m} t+T_{0}^{-3}\right)^{-1 / 3}=T_{0}\left(1+\frac{12 r^{2} \pi \sigma T_{0}^{3}}{c m} t\right)^{-1 / 3}=\frac{16700^{\circ} \mathrm{K}}{\sqrt[3]{1+1.45 \cdot 10^{-16} \mathrm{~s}^{-1} t}}
$$

From the relationship, it can be seen that the temperature of the star

$$
4.8 \cdot 10^{16} \mathrm{~s} \approx 1.5 \text { billion years }
$$

decreases to half in.

Gábor Kárpáti (Kaposvár, Táncsics M. Gymnasium, 4th grade)