To solve the problem, we need to determine how much mercury will spill out of the glass flask when both the flask and the mercury are heated from \(0^\circ C\) to \(100^\circ C\). We will use the coefficients of volume expansion for mercury and linear expansion for glass to find the answer.
### Step-by-Step Solution:
1. **Identify Given Values:**
- Initial volume of the flask and mercury, \(V_0 = 1 \, \text{litre}\)
- Coefficient of volume expansion of mercury, \(\beta_{Hg} = 1.82 \times 10^{-4} \, \text{°C}^{-1}\)
- Coefficient of linear expansion of glass, \(\alpha_{glass} = 0.1 \times 10^{-4} \, \text{°C}^{-1}\)
- Temperature change, \(\Delta T = 100^\circ C - 0^\circ C = 100^\circ C\)
2. **Calculate the Volume Expansion of Mercury:**
The volume expansion of mercury can be calculated using the formula:
\[
V_{Hg} = V_0 (1 + \beta_{Hg} \Delta T)
\]
Substituting the values:
\[
V_{Hg} = 1 \, \text{litre} \times \left(1 + 1.82 \times 10^{-4} \times 100\right)
\]
\[
V_{Hg} = 1 \, \text{litre} \times (1 + 0.0182) = 1 \, \text{litre} \times 1.0182 = 1.0182 \, \text{litres}
\]
3. **Calculate the Volume Expansion of the Glass Flask:**
The volume expansion of the glass flask is given by:
\[
V_{glass} = V_0 (1 + \beta_{glass} \Delta T)
\]
The volumetric expansion coefficient of glass (\(\beta_{glass}\)) can be calculated from the linear expansion coefficient:
\[
\beta_{glass} = 3 \alpha_{glass} = 3 \times 0.1 \times 10^{-4} = 0.3 \times 10^{-4} \, \text{°C}^{-1}
\]
Now substituting the values:
\[
V_{glass} = 1 \, \text{litre} \times \left(1 + 0.3 \times 10^{-4} \times 100\right)
\]
\[
V_{glass} = 1 \, \text{litre} \times (1 + 0.003) = 1 \, \text{litre} \times 1.003 = 1.003 \, \text{litres}
\]
4. **Calculate the Volume of Mercury that Spills Out:**
The volume of mercury that spills out can be found by subtracting the expanded volume of the glass flask from the expanded volume of mercury:
\[
V_{spill} = V_{Hg} - V_{glass}
\]
\[
V_{spill} = 1.0182 \, \text{litres} - 1.003 \, \text{litres} = 0.0152 \, \text{litres}
\]
### Final Answer:
The volume of mercury that will spill out is approximately \(0.0152 \, \text{litres}\).
