Question

How much heat is required to raise the temperature of 250 ML of water from 20-degree Celcius to 35 degrees Celcius? How much heat is lost by the water as it cools back down to 20 degrees Celcius?

Answer 1

250 ML water= 250 ml * 1 gm/ml water= 250 grams

Specific heat of water, c= 4.18 J/g ^oC

To raise the temperature from 20-degree Celsius to 35-degree Celsius (temperature difference, (35-20)=15-degree Celsius), heat required:

Q= mass*specific heat*temperature difference= 250*4.18*15= 15675J.

Again, heat lost by water as it cools back to 20-degree Celsius is:

E=250*4.18*(20-35)= -15675J.