In an alcohol in glass thermometer, the alcohol column has length 12.45 cm at `0.0^(@)C` and length `21.30` cm at `100.0^(@)C`. What is the temperature if the column has length a. 15.10 cm, and b. 22.95 cm.

Answer

To solve the problem, we will follow these steps:

### Given Data:
- Length of alcohol at 0.0°C, \( L_0 = 12.45 \, \text{cm} \)
- Length of alcohol at 100.0°C, \( L_{100} = 21.30 \, \text{cm} \)

### Step 1: Calculate the temperature per unit length
We need to find the change in temperature per unit change in length of the alcohol column.

1. **Change in temperature, \( \Delta T \)**:
\[ \Delta T = 100.0°C - 0.0°C = 100.0°C \]

2. **Change in length, \( \Delta L \)**:
\[ \Delta L = L_{100} - L_0 = 21.30 \, \text{cm} - 12.45 \, \text{cm} = 8.85 \, \text{cm} \]

3. **Temperature per unit length**:
\[ \frac{\Delta T}{\Delta L} = \frac{100.0°C}{8.85 \, \text{cm}} \approx 11.30 \, \frac{°C}{\text{cm}} \]

### Step 2: Calculate the temperature for \( L = 15.10 \, \text{cm} \)
1. **Change in length from reference point**:
\[ L_{15.10} - L_0 = 15.10 \, \text{cm} - 12.45 \, \text{cm} = 2.65 \, \text{cm} \]

2. **Change in temperature for this length**:
\[ \Delta T_{15.10} = \frac{\Delta T}{\Delta L} \times (L_{15.10} - L_0) = 11.30 \, \frac{°C}{\text{cm}} \times 2.65 \, \text{cm} \approx 29.94°C \]

3. **Final temperature**:
\[ T_{15.10} = 0.0°C + 29.94°C \approx 29.94°C \]

### Step 3: Calculate the temperature for \( L = 22.95 \, \text{cm} \)
1. **Change in length from reference point**:
\[ L_{22.95} - L_0 = 22.95 \, \text{cm} - 12.45 \, \text{cm} = 10.50 \, \text{cm} \]

2. **Change in temperature for this length**:
\[ \Delta T_{22.95} = \frac{\Delta T}{\Delta L} \times (L_{22.95} - L_0) = 11.30 \, \frac{°C}{\text{cm}} \times 10.50 \, \text{cm} \approx 118.65°C \]

3. **Final temperature**:
\[ T_{22.95} = 0.0°C + 118.65°C \approx 118.65°C \]

### Final Results:
- a. The temperature corresponding to a length of 15.10 cm is approximately **29.94°C**.
- b. The temperature corresponding to a length of 22.95 cm is approximately **118.65°C**.