A liquid having mass `m = 250` g is kept warm in a vessel by use of an electric heater. The liquid is maintained at `50^(@)C` when the power supplied by heater is 30 watt and surrounding temperature is `20^(@)C`. As the heater is switched off, the liquid starts cooling and it was observed that it took 10 second for temperature to fall down from `40^(@)C` to `39.9^(@)C`. Calculate the specific heat capacity of the liquid. Assume Newton’s law of cooling to be applicable.

Answer

`8000(J)/(kg^(@)C)`