To determine the volume of mercury that overflows as the temperature increases from 20 °C to 50 °C, we first need to calculate the change in volume of both the mercury and the glass flask due to thermal expansion.
Step 1: Calculate the change in volume of mercury
The volumetric expansion formula is given by:
\[ \Delta V_{\text{mercury}} = V_0 \cdot \beta \cdot \Delta T \]
Where:
- \( \Delta V_{\text{mercury}} \) is the change in volume of mercury.
- \( V_0 = 35 , \text{cm}^3 \) (initial volume of mercury).
- \( \beta = 17 \times 10^{-5} , \text{°C}^{-1} \) (volumetric expansion coefficient of mercury).
- \( \Delta T = 50 , \text{°C} - 20 , \text{°C} = 30 , \text{°C} \) (temperature change).
Substituting the values:
\[ \Delta V_{\text{mercury}} = 35 , \text{cm}^3 \cdot (17 \times 10^{-5} , \text{°C}^{-1}) \cdot 30 , \text{°C} \]
Calculating the change:
\[ \Delta V_{\text{mercury}} = 35 \cdot 17 \cdot 30 \times 10^{-5} \]
\[ \Delta V_{\text{mercury}} = 35 \cdot 510 \times 10^{-5} = 17850 \times 10^{-5} = 1.785 , \text{cm}^3 \]
Step 2: Calculate the change in volume of the glass flask
The change in volume of the glass flask can be calculated using the linear expansion coefficient:
\[ \Delta V_{\text{glass}} = V_0 \cdot \alpha \cdot \Delta T \]
Where:
- \( \alpha = 9 \times 10^{-6} , \text{°C}^{-1} \) (linear expansion coefficient of glass).
- The volume of the glass flask increases due to the linear expansion, but since this is a volume expansion, we need to adjust it accordingly. The volumetric expansion for a solid is approximately three times the linear expansion for small changes, thus:
\[ \beta_{\text{glass}} \approx 3 \alpha = 3 \times (9 \times 10^{-6}) = 27 \times 10^{-6} , \text{°C}^{-1} \]
Now we can find the change in volume for the glass:
\[ \Delta V_{\text{glass}} = V_0 \cdot \beta_{\text{glass}} \cdot \Delta T \] Assuming the initial volume of the flask is equal to the volume of mercury:
\[ \Delta V_{\text{glass}} = 35 , \text{cm}^3 \cdot (27 \times 10^{-6} , \text{°C}^{-1}) \cdot 30 , \text{°C} \]
Calculating the change:
\[ \Delta V_{\text{glass}} = 35 \cdot 27 \cdot 30 \times 10^{-6} \]
\[ \Delta V_{\text{glass}} = 35 \cdot 810 \times 10^{-6} = 28350 \times 10^{-6} = 0.02835 , \text{cm}^3 \]
Step 3: Calculate the net volume of mercury that overflows
To find the volume of mercury that overflows, we subtract the change in volume of the glass from the change in volume of the mercury:
\[ V_{\text{overflow}} = \Delta V_{\text{mercury}} - \Delta V_{\text{glass}} \]
Substituting the calculated values:
\[ V_{\text{overflow}} = 1.785 , \text{cm}^3 - 0.02835 , \text{cm}^3 \]
\[ V_{\text{overflow}} \approx 1.75665 , \text{cm}^3 \]
Conclusion
The volume of mercury that overflows is approximately:
\[ \boxed{1.76 , \text{cm}^3} \]