A refrigerator converts 100 g of water at `20^@`C to ice at `- 10^@`C in 73.5 min. Calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 J `g^(-1) K^(-1)`, specific latent heat of ice is 336 J `g^(-1)` and the specific heat capacity of ice is 2.1 J `g^(-1) K^(-1)`

Explanation

To solve the problem of calculating the average rate of heat extraction by the refrigerator, we will follow these steps:

### Step 1: Calculate the heat required to cool water from 20°C to 0°C
The formula to calculate the heat (Q) is:
\[ Q = m \cdot c \cdot \Delta T \]
Where:
- \( m \) = mass of water = 100 g
- \( c \) = specific heat capacity of water = 4.2 J/g°C
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