A tank contains $m$ liters of water at $t \mathrm{C}^{\circ}$. We drain $v$ liters of water from it, and replace it with $v$ liters of water at $d \mathrm{C}^{\circ}$, then we drain $v$ liters of water again, and replace it with the same amount of water at $d \mathrm{C}^{\circ}$, and so on.

a) What will be the temperature of the water in $\mathrm{C}^{\circ}$ if $m=40001, t=60 \mathrm{C}^{\circ}, d=10 \mathrm{C}^{\circ}, v=201$, and the above operation is performed 100 times.

b) How many such water exchanges are needed for the water to cool down to $28 \mathrm{C}^{\circ}$ if $m=40001, t=40 \mathrm{C}^{\circ}, d=15 \mathrm{C}^{\circ}$ and $v=201$.
First, we solve the problem in general. We start from the known relationship that the amount of heat released is equal to the amount of heat absorbed, neglecting heat loss. In this case, therefore, the temperature $t_{1}$ of the water after the first pouring can be calculated from the following equation:

$$
(m-v)\left(t-t_{1}\right)=v\left(t_{1}-d\right)
$$

from which

$$
t_{1}=\frac{(m-v) t+v d}{m}=\frac{(m-v) t+d[m-(m-v)]}{m}
$$

Based on this, the temperature of the water after the second pouring is

$$
\begin{aligned}
t_{2}=\frac{(m-v) t_{1}+v d}{m} & =\frac{(m-v)^{2} t+d\left[m(m-v)-(m-v)^{2}\right]}{m^{2}}+\frac{m v d}{m^{2}}= \\
& =\frac{(m-v)^{2} t+d\left[m^{2}-(m-v)^{2}\right]}{m^{2}}
\end{aligned}
$$

We will prove by induction that after the $n$-th pouring,

$$
t_{n}=\frac{(m-v)^{n} t+d\left[m^{n}-(m-v)^{n}\right]}{m^{n}}
$$

Let

$$
t_{k}=\frac{(m-v)^{k} t+d\left[m^{k}-(m-v)^{k}\right]}{m^{k}}
$$

then, according to (1),

$$
\begin{gathered}
t_{k+1}=\frac{(m-v) t_{k}+v d}{m}=\frac{m-v}{m} t_{k}+\frac{v d}{m}=\frac{(m-v)^{k+1} t+d\left[m^{k}(m-v)-(m-v)^{k+1}\right.}{m^{k+1}}+ \\
+\frac{m^{k} v d}{m^{k+1}}=\frac{(m-v)^{k+1} t+d\left[m^{k+1}-(m-v)^{k+1}\right]}{m^{k+1}}
\end{gathered}
$$

Since we have seen that (2) is true for $n=2$ (and $n=1$), (2) is true for any natural number $n$.

After this, we can apply formula (2) to the current problem.
a) $t_{100}=\frac{3980^{100} \cdot 60+10\left(4000^{100}-3980^{100}\right)}{4000^{100}}=\left(\frac{398}{400}\right)^{100} \cdot 50+10=30.13+10=40.13 \mathrm{C}^{\circ}$

b) Let the number of water changes required be $x$.

$$
t_{x}=28=\frac{3980^{x} 40+15\left(4000^{x}-3980^{x}\right)}{4000^{x}}=\left(\frac{398}{400}\right)^{x} \cdot 25+15
$$

from which

$$
\left(\frac{398}{400}\right)^{x}=\frac{13}{25}
$$

thus

$$
x=\frac{\lg 25-\lg 13}{\lg 400-\lg 398} \sim \frac{0.284}{0.0022}=129.1
$$

Balint Pödör (Bp. II., Rakoczi g. III. o. t.)