The water of mass 75 g at `100^(@)C` is added to ice of mass `20 g` at `-15^(@)C` . What is the resulting temperature. Latent heat of ice = `80 cal//g` and specific heat of ice = `0.5`.

Explanation

To find the resulting temperature when 75 g of water at 100°C is mixed with 20 g of ice at -15°C, we will use the principle of conservation of energy. The heat lost by the water will be equal to the heat gained by the ice.

### Step 1: Define the variables
- Mass of water (M1) = 75 g
- Initial temperature of water (T1) = 100°C
- Specific heat of water (S1) = 1 cal/g°C
- Mass of ice (M2) = 20 g
- Initial temperature of ice (T2) = -15°C
...