## Task 5 - 050835

Someone mixes $9 \mathrm{~kg}$ of water at a temperature of $30^{\circ} \mathrm{C}$ with $6 \mathrm{~kg}$ of water at a temperature of $85^{\circ} \mathrm{C}$ and stirs the mixture well.

What temperature would the mixture assume if heat exchange with the environment is neglected?
If the mass of the first component is denoted by $m_{1}$, that of the second by $m_{2}$, the temperature of the first component by $t_{1}$, and that of the second by $t_{2}$, then for the mixture temperature $t_{x}$ under the given conditions, the following applies:

$$
t_{x}=\frac{m_{1}}{m_{1}+m_{2}} t_{1}+\frac{m_{2}}{m_{1}+m_{2}} t_{2}
$$

Using the given values, we thus obtain, if $x$ represents the value of $t_{x}$,

$$
\begin{aligned}
x & =\frac{9}{15} \cdot 30+\frac{6}{15} \cdot 85, \quad \text { so } \\
x & =52
\end{aligned}
$$

The mixture has a temperature of $t_{x}=52^{\circ} \mathrm{C}$.