`1 kg` ice at `-20^(@)C` is mixed with `1 kg` steam at `200^(@)C`. The equilibrium temperature and mixture content is

Explanation

To solve the problem of mixing `1 kg` of ice at `-20°C` with `1 kg` of steam at `200°C`, we need to calculate the equilibrium temperature and the mixture content step by step.

### Step 1: Calculate the heat required to convert ice at `-20°C` to water at `100°C`.

1. **Heating the ice from `-20°C` to `0°C`:**
- Mass of ice, \( m = 1 \, \text{kg} = 1000 \, \text{g} \)
- Specific heat of ice, \( s_{ice} = 0.5 \, \text{cal/g°C} \)
- Temperature change, \( \Delta T_1 = 0 - (-20) = 20 \, \text{°C} \)
...