Calculate the heat of neutralisation from the following data:
`200mL` of `1M HCI` is mixed with `400mL` of `0.5M NaOH`. The temperature rise in calorimeter was found to be `4.4^(@)C`. Water equivalent of calorimeter is `12g` and specific heat is `1cal mL^(-1) degree^(-1)` for solution.

Explanation

To calculate the heat of neutralization from the given data, we can follow these steps:

### Step 1: Calculate the moles of HCl and NaOH
- **Moles of HCl**:
\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 1 \, \text{mol/L} \times 0.200 \, \text{L} = 0.2 \, \text{mol} \]

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