200 g of water at `50.5^@C` is cooled down to `10^@C` by adding m g of ice cubes at `0^@C` in it. Find m. Take, specific heat capacity of water = `4.2Jg^(-1)""^@C^(-1)` and specific latent heat of ice = `336Jg^(-1)`

Explanation

To solve the problem of finding the mass of ice cubes (m) that needs to be added to cool down 200 g of water from 50.5°C to 10°C, we will follow these steps:

### Step 1: Identify the given data
- Mass of water (M) = 200 g
- Initial temperature of water (Ti) = 50.5°C
- Final temperature of water (Tf) = 10°C
- Specific heat capacity of water (c) = 4.2 J/g°C
- Specific latent heat of ice (L) = 336 J/g
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