40 g of ice at `0^(@)C` is used to bring down the temperature of a certain mass of water at `60^(@)C` to `10^(@)С`. Find the mass of water used.
[Specific heat capacity of water = 4200 `Jkg^(-1)""^(@)C^(-1)`]
[Specific latent heat of fusion of ice = `336xx10^(3)Jkg^(-1)`]

Explanation

To solve the problem, we will use the principle of conservation of energy, which states that the heat lost by the water will be equal to the heat gained by the ice.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of ice, \( m_2 = 40 \, \text{g} = 0.04 \, \text{kg} \) (since 1 g = 0.001 kg)
- Initial temperature of ice, \( T_{i2} = 0^\circ C \)
- Final temperature of water, \( T_f = 10^\circ C \)
...