Water (2330 g) is heated until it just begins to boil. If the water absorbs 5.53×105 J of heat in the process, what was the initial temperature of the water?

Answer 1

Q=mc∆t
Q = heat energy
m = mass
c = specific heat (of water)
∆t= change in temperature (= final - initial temp)

final temp= boiling point of water i.e. 100*C
c= 4.186 joule/gram °C

Answer 2

This is how far i got
2330g *4.184J/g degree C *(100 degree C-unk) =537,000J

so

9748.72J * unk =537,000J

so

537000/9748.72=55.08

so

100-55.08= Initial temp of 44.92 degrees celsius

:)