30.4986815-4.97*10^5

Question

Water (3110g ) is heated until it just begins to boil. If the water absorbs 4.97 × 10^5 J of heat in the process, what was the initial temperature of the water?

Answer 1

30.4986815-4.97×10⁵=

30.4986815-497000=

-496969.5013185.

 

3110g=3.11kg.

Specific heat of water=4200J/kg/℃.

So 3.11kg requires 13062J to raise the temperature 1℃.

4.97×10⁵=497000J. This is enough energy to raise the water 497000/13062=38.05℃. Since water boils at 100℃, the temperature of the water must have been 100-38.05=61.95℃.

This assumes that the water reaches boiling point, but does not actually boil, so we don’t have to consider the latent heat of vaporisation.