0.5 kg of lemon squash at `30^@`C is placed in a refrigerator which can remove heat at an average rate of 30 J `s^(-1)`. How long will it take to cool the lemon squash to `5^@`C ? Specific heat capacity of squash = 4200 J `kg^(-1) K^(-1)`.

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given values
- Mass of lemon squash, \( m = 0.5 \, \text{kg} \)
- Initial temperature, \( T_i = 30 \, ^\circ C \)
- Final temperature, \( T_f = 5 \, ^\circ C \)
- Specific heat capacity, \( s = 4200 \, \text{J kg}^{-1} \text{K}^{-1} \)
- Rate of heat removal by the refrigerator, \( R = 30 \, \text{J s}^{-1} \)
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