250 g of water at `30^(@)C` is present in a copper vessel of mass 50 g. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to `5^(@)C`.
Specific latent heat of fusion of ice = `336xx10^(3)Jkg^(-1)`
Specific heat capacity of copper vessel = `400Jkg^(-1)""^(@)C^(-1)`
Specific heat capacity of water = `4200Jkg^(-1)""^(@)C^(-1)`.

Explanation

To solve the problem, we will use the principle of calorimetry, which states that the heat lost by the water and the copper vessel will be equal to the heat gained by the ice as it melts and warms up to the final temperature.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of water, \( m_w = 250 \, \text{g} = 0.25 \, \text{kg} \)
- Initial temperature of water, \( T_{w_i} = 30^\circ C \)
- Mass of copper vessel, \( m_c = 50 \, \text{g} = 0.05 \, \text{kg} \)
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