10 gm of ice cubes at `0^(@)"C"` are released in a tumbler (water equivalent 55 g) at `40^(@)"C"`. Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearly` "(L = 80 cal/g)"`

Options

A. `31^(@)C`
B. `22^(@)C`
C. `19^(@)C`
D. `15^(@)C`

Explanation

To solve the problem, we need to find the final equilibrium temperature (T2) when 10 grams of ice at 0°C is added to a tumbler containing water (with a water equivalent of 55 grams) at 40°C. We will use the principle of conservation of energy, where the heat lost by the water will equal the heat gained by the ice.

### Step-by-step Solution:

1. **Calculate the heat required to melt the ice:**
The heat required to convert ice at 0°C to water at 0°C can be calculated using the formula:
\[ Q_{\text{ice}} = m_{\text{ice}} \times L_f \]
where:
- \( m_{\text{ice}} = 10 \, \text{g} \)
- \( L_f = 80 \, \text{cal/g} \)

Substituting the values:
\[ Q_{\text{ice}} = 10 \, \text{g} \times 80 \, \text{cal/g} = 800 \, \text{cal} \]

2. **Calculate the heat lost by the water:**
The heat lost by the water when it cools from 40°C to the final temperature \( T_2 \) can be calculated using:
\[ Q_{\text{water}} = m_{\text{water}} \times c \times (T_{\text{initial}} - T_2) \]
where:
- \( m_{\text{water}} = 55 \, \text{g} \)
- \( c = 1 \, \text{cal/g°C} \)
- \( T_{\text{initial}} = 40 \, \text{°C} \)

Thus:
\[ Q_{\text{water}} = 55 \, \text{g} \times 1 \, \text{cal/g°C} \times (40 - T_2) \]
\[ Q_{\text{water}} = 55 \times (40 - T_2) \, \text{cal} \]

3. **Set the heat gained by the ice equal to the heat lost by the water:**
According to the conservation of energy:
\[ Q_{\text{ice}} = Q_{\text{water}} \]
Therefore:
\[ 800 = 55 \times (40 - T_2) \]

4. **Solve for \( T_2 \):**
Rearranging the equation:
\[ 800 = 2200 - 55T_2 \]
\[ 55T_2 = 2200 - 800 \]
\[ 55T_2 = 1400 \]
\[ T_2 = \frac{1400}{55} \approx 25.45 \, \text{°C} \]

5. **Final Calculation:**
Since the problem states that the final temperature is nearly \( T_2 \), we can round it to the nearest whole number:
\[ T_2 \approx 25 \, \text{°C} \]

### Conclusion:
The final temperature of the water in the tumbler after adding the ice is approximately **25°C**.