Is it possible to use $1 \mathrm{~kg}$ of water at $100{ }^{\circ} \mathrm{C}$ to heat $1 \mathrm{~kg}$ of water at $0{ }^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$? We can neglect the heat capacity of the containers. Estimate the highest achievable temperature!
We need to heat the $0{ }^{\circ} \mathrm{C}$ water at the expense of cooling the $100{ }^{\circ} \mathrm{C}$ water; the more the originally $100{ }^{\circ} \mathrm{C}$ water cools, the hotter the $0{ }^{\circ} \mathrm{C}$ water will be at the end of the process. Our goal is to cool the $100{ }^{\circ} \mathrm{C}$ water as much as possible, but we must always keep in mind that the direction of heat flow is clearly defined; the hotter body transfers heat to the colder body.

If we bring only a very small part of the $100{ }^{\circ} \mathrm{C}$ water into thermal contact with the $0{ }^{\circ} \mathrm{C}$ water, then it will cool down very strongly, and thus, from the perspective of heating, its utilization is very good. It is worth trying to divide the hot water into small parts and heat the colder water separately with each part.

Divide the $100{ }^{\circ} \mathrm{C}$ water into $n$ equal parts. Connect the first part with a very good, lossless heat conductor to the $0{ }^{\circ} \mathrm{C}$ water. The equilibrium temperature $T_{1}$ can be determined from the equation

$$
c m\left(T_{1}-T_{0}\right)=c(m / n)\left(100{ }^{\circ} \mathrm{C}-T_{1}\right)
$$

($m=1 \mathrm{~kg}, T_{0}=0{ }^{\circ} \mathrm{C}, c$ is the specific heat capacity of water):

$$
T_{1}=100{ }^{\circ} \mathrm{C} /(n+1)
$$

If $n$ is large enough, then $T_{1}$ is quite small, the $100{ }^{\circ} \mathrm{C}$ water cools well during the heat exchange, that is, we have utilized it very well.

Then connect the good heat conductor to the $T_{1}$ temperature water and the second portion of $100{ }^{\circ} \mathrm{C}$ water. The equilibrium temperature $T_{2}$ is:

$$
T_{2}=\frac{100{ }^{\circ} \mathrm{C}}{n+1}+\frac{n \cdot 100^{\circ} \mathrm{C}}{(n+1)^{2}}
$$

$T_{2}>T_{1}$, so the second portion of our $100{ }^{\circ} \mathrm{C}$ water has cooled less than the first, and thus we have utilized it less for heating. It is worth writing down the common temperature after heating with the third portion:

$$
\begin{aligned}
T_{3}=\frac{100{ }^{\circ} \mathrm{C}}{n+1}+ & \frac{n}{n+1} T_{2}=\frac{100{ }^{\circ} \mathrm{C}}{n+1}+\frac{n}{n+1} \cdot \frac{100{ }^{\circ} \mathrm{C}}{n+1}+\frac{n^{2} \cdot 100{ }^{\circ} \mathrm{C}}{(n+1)^{3}}= \\
& =\frac{100}{n+1}\left(1+\frac{n}{n+1}+\frac{n^{2}}{(n+1)^{2}}\right){ }^{\circ} \mathrm{C}
\end{aligned}
$$

It is clearly visible that the resulting common temperature is determined by the sum of a geometric series, so the final temperature is

$$
T_{n}=\frac{100{ }^{\circ} \mathrm{C}}{n+1} \cdot\left[\frac{\left(\frac{n}{n+1}\right)^{n}-1}{\frac{n}{n+1}-1}\right]=100{ }^{\circ} \mathrm{C}\left[1-\left(\frac{n}{n+1}\right)^{n}\right]
$$

$[n /(n+1)]^{n}$ is a well-known expression whose reciprocal approaches the base of the natural logarithm, $e \approx 2.72$, as $n \rightarrow \infty$. Therefore, if we divide the $100{ }^{\circ} \mathrm{C}$ water into a very large number of parts, we can heat the originally $0{ }^{\circ} \mathrm{C}$ water to $T \approx 63.2{ }^{\circ} \mathrm{C}$ using the above procedure.