`20 g` of steam at `100^@ C` is passes into `100 g` of ice at `0^@C`. Find the resultant temperature if latent heat if steam is `540 cal//g`,latent heat of ice is `80 cal//g` and specific heat of water is `1 cal//g^@C`.

Explanation

To solve the problem, we need to calculate the heat released by the steam as it condenses into water and cools down to the final temperature \( T \), and the heat absorbed by the ice as it melts and warms up to the same final temperature \( T \).

### Step-by-Step Solution:

1. **Calculate the heat released by steam as it condenses to water:**
- The mass of steam = \( 20 \, \text{g} \)
- Latent heat of steam = \( 540 \, \text{cal/g} \)
- Heat released during condensation:
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