What amount of ice at `-14^(@)C` required to cool `200gg` of water from `25^(@)C` to 10CC?
(Given `C_("ice")=0.5cal/g.^(@)C,L_(f)` for ice `=80cal//g`)

Options

A. `31g`
B. `41g`
C. `51g`
D. `21g`

Explanation

To solve the problem of how much ice at -14°C is required to cool 200 grams of water from 25°C to 10°C, we will follow these steps:

### Step 1: Calculate the heat lost by the water
The heat lost by the water can be calculated using the formula:
\[ Q = m \cdot C \cdot \Delta T \]
Where:
- \( m \) = mass of water = 200 g
- \( C \) = specific heat of water = 1 cal/g°C (standard value)
...