100 gm of water at `20^@C` is converted into ice at `0^@C` by a refrigerator in 2 hours. What will be the quantity of heat removed per minute? Specific heat of water = `1 cal g^(-1) C^(-1) ` and latent heat of ice `= 80 cal g^(-1)`

Explanation

To find the quantity of heat removed per minute when 100 grams of water at 20°C is converted into ice at 0°C, we can follow these steps:

### Step 1: Calculate the heat removed to cool the water from 20°C to 0°C.

The formula to calculate the heat removed when cooling is given by:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \( Q \) = heat removed (in calories)
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