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 in an insulated vessel, we can follow these steps:

### Step 1: Understand the Initial Conditions
We have:
- 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, \( T_i = 253 \, \text{K} \)

### Step 2: Calculate Heat Loss from Steam
When steam condenses to water at 0°C (273 K), it loses heat. The heat loss can be calculated using the formula:
\[ Q_s = m_s \cdot L_v + m_s \cdot s_w \cdot (T_s - T_f) \]
Where:
- \( L_v = 2268 \, \text{J/g} \) (latent heat of vaporization)
- \( s_w = 4200 \, \text{J/kg-K} = 4.2 \, \text{J/g-K} \) (specific heat of water)
- \( T_f \) is the final temperature (which we will assume to be 273 K for now).

Calculating the heat loss:
1. Convert the mass of steam to grams: \( m_s = 50 \, \text{g} \)
2. Calculate the heat lost during condensation:
\[ Q_{condensation} = m_s \cdot L_v = 50 \, \text{g} \cdot 2268 \, \text{J/g} = 113400 \, \text{J} \]
3. Calculate the heat lost when cooling from 373 K to 273 K:
\[ Q_{cooling} = m_s \cdot s_w \cdot (T_s - T_f) = 50 \, \text{g} \cdot 4.2 \, \text{J/g-K} \cdot (373 - 273) = 50 \cdot 4.2 \cdot 100 = 21000 \, \text{J} \]
4. Total heat loss from steam:
\[ Q_s = Q_{condensation} + Q_{cooling} = 113400 \, \text{J} + 21000 \, \text{J} = 134400 \, \text{J} \]

### Step 3: Calculate Heat Required to Melt Ice
The heat required to convert ice at 253 K to water at 273 K can be calculated as follows:
\[ Q_i = m_i \cdot L_f + m_i \cdot s_i \cdot (T_f - T_i) \]
Where:
- \( L_f = 336 \, \text{J/g} \) (latent heat of fusion)
- \( s_i = 2100 \, \text{J/kg-K} = 2.1 \, \text{J/g-K} \) (specific heat of ice)

Calculating the heat required:
1. Calculate the heat required for melting:
\[ Q_{melting} = m_i \cdot L_f = 450 \, \text{g} \cdot 336 \, \text{J/g} = 151200 \, \text{J} \]
2. Calculate the heat required to raise the temperature of melted ice from 253 K to 273 K:
\[ Q_{heating} = m_i \cdot s_w \cdot (T_f - T_i) = 450 \, \text{g} \cdot 4.2 \, \text{J/g-K} \cdot (273 - 253) = 450 \cdot 4.2 \cdot 20 = 37800 \, \text{J} \]
3. Total heat required for ice:
\[ Q_i = Q_{melting} + Q_{heating} = 151200 \, \text{J} + 37800 \, \text{J} = 189000 \, \text{J} \]

### Step 4: Compare Heat Loss and Heat Required
- Heat lost by steam: \( Q_s = 134400 \, \text{J} \)
- Heat required by ice: \( Q_i = 189000 \, \text{J} \)

Since the heat required to melt the ice and raise its temperature is greater than the heat lost by the steam, not all the ice will melt. The final temperature will be the melting point of ice, which is 273 K.

### Final Answer
The final temperature of the mixture is:
\[ \boxed{273 \, \text{K}} \]