An electric heater of power 1000 W raises the temperature of 5 kg of a liquid from `25^@`C to `31^@`C in 2 minutes. Calculate : (i) the heat capacity, and (ii) the specific heat capacity of liquid.

Explanation

To solve the problem step by step, we will calculate both the heat capacity and the specific heat capacity of the liquid.

### Given Data:
- Power of the heater (P) = 1000 W
- Mass of the liquid (m) = 5 kg
- Initial temperature (T1) = 25 °C
- Final temperature (T2) = 31 °C
- Time (t) = 2 minutes = 2 × 60 seconds = 120 seconds
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