In an experiment, it took 300s to increase the temperature of 0.1kg of the liquid from 25°c to 50°c. during this period, the energy supplied by the heater was 13600j, and the apparatus was losing heat at an average rate of 12j/s. assuming the heat capacity of the container can be ignored, calculate the value of the specific heat capacity of the liquid. pls i need urgent !

Explanation:

        The temperature of 0.1 kg of liquid rises from 25°C to 50°C in 300 sec. Energy of 13,600 J was supplied during this time. Appartus was losing energy at the rate of 12 J/sec.

       Let us assume the Specific heat capacity as .

      As there is no state change from liquid to gas, only Specific heat capacity is involved. Also, work done is approximately zero because volume does not change much. So,

       Energy gained = Energy required to rise the temperature

       Energy gained by liquid =

∴ Specific heat capacity of liquid = 4000

answer: the answer is a

the answer is b, have negligible mass