A calorimeter of mass 60 g contains 180 g of water at `29^(@)C`. Find the final temperature of the mixture, when 37.2 g of ice at -`10^(@)C` is added to it (specific heat capacity of water = `4200 J//kg `K. latent heat of ice `336 xx 10^(3)J//kg`. Specific heat capacity of ice `= 2100 J//kg K`, specific heat capacity of the calorimeter is `0.42 Jg^(-1) xx C^(-1)`)

Explanation

To find the final temperature of the mixture when ice is added to the water in the calorimeter, we can use the principle of conservation of energy. The heat lost by the water and the calorimeter will be equal to the heat gained by the ice.

### Step-by-Step Solution:

1. **Identify the given data:**
- Mass of the calorimeter (m_cal) = 60 g = 0.060 kg
- Mass of water (m_water) = 180 g = 0.180 kg
- Initial temperature of water (T_initial_water) = 29 °C
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