A vessel contains `110 g` of water. The heat capacity of the vessel is equal to `10 g` of water. The initial temperature of water in vessel is `10^(@)C`. If `220 g` of hot water at `70^(@)C` is poured in the vessel, the final temperature neglecting radiation loss, will be

Options

A. `70^(@)C`
B. `80^(@)C`
C. `60^(@)C`
D. `50^(@)C`

Explanation

To find the final temperature of the water in the vessel after adding hot water, we can use the principle of conservation of energy, which states that the heat lost by the hot water will be equal to the heat gained by the cold water and the vessel.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of cold water in the vessel, \( m_1 = 110 \, \text{g} \)
- Initial temperature of cold water, \( T_1 = 10^\circ C \)
- Heat capacity of the vessel (equivalent to water), \( m_v = 10 \, \text{g} \)
- Mass of hot water added, \( m_2 = 220 \, \text{g} \)
- Initial temperature of hot water, \( T_2 = 70^\circ C \)

2. **Let the Final Temperature be \( \theta \):**
- The final temperature of the mixture will be \( \theta \).

3. **Calculate Heat Gained by Cold Water and Vessel:**
- Heat gained by the cold water:
\[ Q_{\text{cold}} = m_1 \cdot c \cdot (T_f - T_1) + m_v \cdot c \cdot (T_f - T_1) \]
where \( c \) is the specific heat capacity of water (approximately \( 1 \, \text{cal/g}^\circ C \)).
\[ Q_{\text{cold}} = 110 \cdot 1 \cdot (\theta - 10) + 10 \cdot 1 \cdot (\theta - 10) \]
\[ Q_{\text{cold}} = 120(\theta - 10) \]

4. **Calculate Heat Lost by Hot Water:**
- Heat lost by the hot water:
\[ Q_{\text{hot}} = m_2 \cdot c \cdot (T_2 - T_f) \]
\[ Q_{\text{hot}} = 220 \cdot 1 \cdot (70 - \theta) \]

5. **Set Up the Heat Transfer Equation:**
- According to the conservation of energy:
\[ Q_{\text{cold}} = Q_{\text{hot}} \]
\[ 120(\theta - 10) = 220(70 - \theta) \]

6. **Solve for \( \theta \):**
- Expanding both sides:
\[ 120\theta - 1200 = 15400 - 220\theta \]
- Rearranging the equation:
\[ 120\theta + 220\theta = 15400 + 1200 \]
\[ 340\theta = 16600 \]
\[ \theta = \frac{16600}{340} \approx 48.82^\circ C \]

7. **Final Result:**
- The final temperature \( \theta \approx 49^\circ C \) (approximately \( 50^\circ C \)).