## Task 3 - 010913

To maintain the cutting ability of tools when machining metals, cooling is often done with an emulsion of lubricated mineral oil (density $0.98 \mathrm{~g} / \mathrm{cm}^{3}$) and as soft water as possible (density $1.0 \mathrm{~g} / \mathrm{cm}^{3}$). The mixture must have a density of $0.996 \mathrm{g} / \mathrm{cm}^{3}$ for cutting tools of higher hardness, and a density of $0.992 \mathrm{~g} / \mathrm{cm}^{3}$ for grinding operations. How many liters of lubricated mineral oil and how many liters of soft water are needed for 10 liters of emulsion in each case?
The density $\varrho$ of a substance is the quotient of the mass $m$ and the volume $V$ of a certain amount of this substance. Since we are dealing with a mixture of two substances here, the mixing rule must be observed. According to this, the density $\varrho_{E}$ of the emulsion is given by

$$
\varrho_{E}=\frac{\varrho_{\ddot{O} l} \cdot V_{\ddot{O} l}+\varrho_{W} \cdot V_{W}}{V_{\ddot{O} l}+V_{W}}
$$

where $\varrho_{O ̈ l}, V_{\ddot{O} l}$ are the density and volume of the oil used, and $\varrho_{W}, V_{W}$ are the density and volume of the water used. Since the total volume $V=V_{\ddot{O} l}+V_{W}=10 l=10000 \mathrm{~cm}^{3}$ is given according to the problem statement, we can substitute $V_{\ddot{O} l}=V-V_{W}$ in (1) and obtain after rearranging

$$
V_{W}=\frac{\varrho_{E}-\varrho_{\ddot{O} l}}{\varrho_{W}-\varrho_{\ddot{O} l}} V
$$

If the emulsion is to have a density of $0.996 \mathrm{~g} / \mathrm{cm}^{3}$, then one needs 81 liters of water and $2 \mathrm{l}$ of oil. For an emulsion with a density of $0.992 \mathrm{~g} / \mathrm{cm}^{3}$, one needs $6 \mathrm{l}$ of water and $4 \mathrm{l}$ of oil.