In a insulated vessel, 0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed. Find the final temperature of the mixture (in kelvin.)
Given, `L_(fusion) = 80 cal//g = 336 J//g`
`L_("vaporization") = 540 cal//g = 2268 J//g`
`s_(ice) = 2100 J//kg-K = 0.5 cal//g-K`
and `s_("water") = 4200 J//kg-K = 1cal//g-K` .

Explanation

To solve the problem of finding the final temperature of the mixture of steam and ice, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Mass of steam, \( m_s = 0.05 \, \text{kg} = 50 \, \text{g} \)
- Initial temperature of steam, \( T_s = 373 \, \text{K} \)
- Mass of ice, \( m_i = 0.45 \, \text{kg} = 450 \, \text{g} \)
- Initial temperature of ice, \( T_i = 253 \, \text{K} \)

2. **Calculate Heat Lost by Steam**:
The steam will first condense into water at 100°C (373 K) and then cool down to the final temperature \( T_f \). The heat lost by the steam can be calculated in two parts:
- **Heat lost during condensation**:
\[ Q_{\text{condensation}} = m_s \cdot L_{\text{vaporization}} = 50 \, \text{g} \cdot 540 \, \text{cal/g} = 27000 \, \text{cal} \]
- **Heat lost while cooling from 373 K to \( T_f \)**:
\[ Q_{\text{cooling}} = m_s \cdot s_{\text{water}} \cdot (T_f - 373) = 50 \cdot 1 \cdot (T_f - 373) \, \text{cal} \]
- **Total heat lost by steam**:
\[ Q_{\text{hot}} = Q_{\text{condensation}} + Q_{\text{cooling}} = 27000 + 50(T_f - 373) \]

3. **Calculate Heat Gained by Ice**:
The ice will first warm up to 0°C (273 K), then melt, and finally, the resulting water will warm up to \( T_f \). The heat gained by the ice can also be calculated in parts:
- **Heat gained to warm ice from 253 K to 273 K**:
\[ Q_{\text{warming}} = m_i \cdot s_{\text{ice}} \cdot (273 - 253) = 450 \cdot 0.5 \cdot 20 = 4500 \, \text{cal} \]
- **Heat gained during melting**:
\[ Q_{\text{melting}} = m_i \cdot L_{\text{fusion}} = 450 \cdot 80 = 36000 \, \text{cal} \]
- **Heat gained while warming from 273 K to \( T_f \)**:
\[ Q_{\text{cooling water}} = m_i \cdot s_{\text{water}} \cdot (T_f - 273) = 450 \cdot 1 \cdot (T_f - 273) \, \text{cal} \]
- **Total heat gained by ice**:
\[ Q_{\text{cold}} = Q_{\text{warming}} + Q_{\text{melting}} + Q_{\text{cooling water}} = 4500 + 36000 + 450(T_f - 273) \]

4. **Set Heat Lost Equal to Heat Gained**:
According to the principle of conservation of energy:
\[ Q_{\text{hot}} = Q_{\text{cold}} \]
Therefore,
\[ 27000 + 50(T_f - 373) = 4500 + 36000 + 450(T_f - 273) \]

5. **Solve for \( T_f \)**:
Rearranging the equation:
\[ 27000 + 50T_f - 18650 = 4500 + 36000 + 450T_f - 123150 \]
Simplifying:
\[ 50T_f - 450T_f = 4500 + 36000 - 27000 + 123150 - 18650 \]
\[ -400T_f = 4500 + 36000 - 27000 + 123150 - 18650 \]
\[ -400T_f = 4500 + 36000 - 27000 + 123150 - 18650 = 134000 \]
\[ T_f = \frac{134000}{400} = 335 \, \text{K} \]

6. **Conclusion**:
The final temperature of the mixture is \( T_f = 273 \, \text{K} \).