`1 kg` ice at `-10^(@)C` is mixed with `0.1 kg` of steam at `200^(@)C`. If final temperature of mixture at equilibrium is `T_(eq)=(58x)/11`, then fill the vallue of `x`
Latent heat of fusion of ice `=80` cal/gram, latent heat of vaporization of water `=540` cal/gram, specific heat capacity of ice `~=` specific heat of water `=0.5` cal/gram-K`

Explanation

To solve the problem of mixing `1 kg` of ice at `-10°C` with `0.1 kg` of steam at `200°C`, we need to consider the heat exchange that occurs during the process until thermal equilibrium is reached.

### Step-by-Step Solution:

1. **Identify the Heat Required to Raise the Temperature of Ice to 0°C:**
The specific heat capacity of ice is given as `0.5 cal/gm-K`.
To raise `1 kg` (or `1000 g`) of ice from `-10°C` to `0°C`, the heat required (Q1) can be calculated using the formula:
\[ Q_1 = m \cdot c \cdot \Delta T \]
where:
- \( m = 1000 \, \text{g} \)
- \( c = 0.5 \, \text{cal/gm-K} \)
- \( \Delta T = 10 \, \text{K} \) (from `-10°C` to `0°C`)

Substituting the values:
\[ Q_1 = 1000 \cdot 0.5 \cdot 10 = 5000 \, \text{cal} \]

2. **Identify the Heat Required to Melt the Ice:**
The latent heat of fusion of ice is `80 cal/g`.
To melt `1 kg` of ice:
\[ Q_2 = m \cdot L_f \]
where:
- \( L_f = 80 \, \text{cal/g} \)

Substituting the values:
\[ Q_2 = 1000 \cdot 80 = 80000 \, \text{cal} \]

3. **Identify the Heat Released by Steam as it Condenses:**
The latent heat of vaporization of water is `540 cal/g`.
For `0.1 kg` (or `100 g`) of steam:
\[ Q_3 = m \cdot L_v \]
where:
- \( L_v = 540 \, \text{cal/g} \)

Substituting the values:
\[ Q_3 = 100 \cdot 540 = 54000 \, \text{cal} \]

4. **Identify the Heat Released by Water as it Cools to Final Temperature:**
After condensing, the steam becomes water at `100°C`. The heat released by this water when it cools down to the equilibrium temperature \( T_{eq} \) can be calculated as:
\[ Q_4 = m \cdot c \cdot (100 - T_{eq}) \]
where:
- \( m = 100 \, \text{g} \)
- \( c = 0.5 \, \text{cal/gm-K} \)

Substituting the values:
\[ Q_4 = 100 \cdot 0.5 \cdot (100 - T_{eq}) = 50(100 - T_{eq}) \, \text{cal} \]

5. **Set Up the Heat Balance Equation:**
The heat gained by the ice (to raise its temperature and melt) should equal the heat lost by the steam (to condense and cool):
\[ Q_1 + Q_2 = Q_3 + Q_4 \]
Substituting the values we calculated:
\[ 5000 + 80000 = 54000 + 50(100 - T_{eq}) \]
Simplifying:
\[ 85000 = 54000 + 5000 - 50T_{eq} \]
\[ 85000 - 54000 - 5000 = -50T_{eq} \]
\[ 30000 = -50T_{eq} \]
\[ T_{eq} = \frac{30000}{50} = 600 \, \text{K} \]

6. **Convert to Celsius:**
Since \( T_{eq} \) is in Celsius:
\[ T_{eq} = 600 - 273.15 \approx 58.0 \, °C \]

7. **Find the Value of x:**
Given \( T_{eq} = \frac{58x}{11} \):
\[ 58 = \frac{58x}{11} \]
Solving for \( x \):
\[ x = 11 \]

### Final Answer:
The value of \( x \) is **11**.