In the past it was practice to measure temperature using so called weight thermometer which consists of glass sphere having a narrow flow out tube. Such a certain weight thermometer is completely filled with 14.578 g of liquid at `15^(@)C`. It is found that 1.232 g of liquid over flows when the temperature is raised to `100^(@)C`. How much more liquid overflow when the temperature is further raised to `200^(@)C`?

Answer

To solve the problem, we need to determine how much more liquid overflows when the temperature is raised from \(100^\circ C\) to \(200^\circ C\). We can use the concept of thermal expansion and the relationship between the overflowed mass of liquid and the temperature change.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Initial mass of liquid at \(15^\circ C\) is \(M_1 = 14.578 \, \text{g}\).
- When the temperature is raised to \(100^\circ C\), the mass of liquid that overflows is \(M_{overflow1} = 1.232 \, \text{g}\).

2. **Calculate Temperature Change for First Overflow:**
- The temperature change from \(15^\circ C\) to \(100^\circ C\) is:
\[ \Delta T_1 = 100 - 15 = 85^\circ C \]

3. **Establish the Relationship for Mass Overflow:**
- The mass overflow can be related to the temperature change. We can express this as:
\[ M_{overflow1} = k \cdot \Delta T_1 \]
- Where \(k\) is a constant that relates the mass overflow to the temperature change.

4. **Set Up the Equation for the Second Overflow:**
- Now, we need to find the mass overflow when the temperature is raised from \(100^\circ C\) to \(200^\circ C\):
- The temperature change for the second case is:
\[ \Delta T_2 = 200 - 100 = 100^\circ C \]
- Let \(M_{overflow2}\) be the mass that overflows in this case:
\[ M_{overflow2} = k \cdot \Delta T_2 \]

5. **Relate the Two Cases:**
- From the first case, we can express \(k\):
\[ k = \frac{M_{overflow1}}{\Delta T_1} = \frac{1.232 \, \text{g}}{85^\circ C} \]
- Substitute \(k\) into the equation for \(M_{overflow2}\):
\[ M_{overflow2} = \left(\frac{1.232 \, \text{g}}{85^\circ C}\right) \cdot 100^\circ C \]

6. **Calculate \(M_{overflow2}\):**
- Calculate the value:
\[ M_{overflow2} = \frac{1.232 \times 100}{85} \approx 1.449 \, \text{g} \]

7. **Final Result:**
- Therefore, the mass of liquid that overflows when the temperature is raised from \(100^\circ C\) to \(200^\circ C\) is approximately \(1.449 \, \text{g}\).

### Summary:
When the temperature is raised from \(100^\circ C\) to \(200^\circ C\), approximately \(1.449 \, \text{g}\) of liquid overflows.