4. Hot oil at a temperature of $100^{\circ} \mathrm{C}$ in a volume of two liters is mixed with one liter of cold oil at a temperature of $20^{\circ} \mathrm{C}$. What volume will the mixture have when thermal equilibrium is established in the mixture? Heat losses to the external environment can be neglected. The coefficient of volumetric expansion of the oil is $2 \cdot 10^{-3} 1 /{ }^{\circ} \mathrm{C}$.
Answer: 3

Let $V_{1}=2$ L be the volume of hot oil, and $V_{2}=1$ L be the volume of cold oil. Then we can write $V_{1}=U_{1}\left(1+\beta t_{1}\right), V_{2}=U_{2}\left(1+\beta t_{2}\right)$, where $U_{1}, U_{2}$ are the volumes of the respective portions of oil at zero temperature; $t_{1}=100^{\circ} \mathrm{C}, t_{2}=20^{\circ} \mathrm{C} ; \beta=2 \cdot 10^{-3} 1 /{ }^{\circ} \mathrm{C}$ is the coefficient of volumetric expansion.

From the heat balance equation: $c m_{1}\left(t-t_{1}\right)+c m_{2}\left(t-t_{2}\right)=0 \Leftrightarrow t=\frac{m_{1} t_{1}+m_{2} t_{2}}{m_{1}+m_{2}}$, where $m_{1}, m_{2}$ are the masses of the hot and cold portions of oil, respectively; $t$ is the temperature of the mixture. Note that $m_{1}=U_{1} \rho, m_{2}=U_{2} \rho$, where $\rho$ is the density of the oil at zero temperature.

Then $W_{1}=U_{1}(1+\beta t), W_{2}=U_{2}(1+\beta t)$. Here $W_{1}, W_{2}$ are the volumes of the initial portions of oil at temperature $t$.

The new total volume of the oil will be equal to

$$
\begin{aligned}
& W_{1}+W_{2}=\left(U_{1}+U_{2}\right)(1+\beta t)=\left(\frac{m_{1}}{\rho}+\frac{m_{2}}{\rho}\right)(1+\beta t)=\frac{1}{\rho}\left(m_{1}+m_{2}+\left(m_{1}+m_{2}\right) \beta t\right)= \\
& \frac{1}{\rho}\left(m_{1}+m_{2}+m_{1} \beta t_{1}+m_{2} \beta t_{2}\right)=\frac{1}{\rho}\left(m_{1}+m_{1} \beta t_{1}\right)+\frac{1}{\rho}\left(m_{2}+m_{2} \beta t_{2}\right)=\frac{m_{1}}{\rho}\left(1+\beta t_{1}\right)+\frac{m_{2}}{\rho}(1+
\end{aligned}
$$

$\left.\beta t_{2}\right)$

That is, $W_{1}+W_{2}=U_{1}\left(1+\beta t_{1}\right)+U_{2}\left(1+\beta t_{2}\right)=V_{1}+V_{2}$

Thus, the volume will not change.