If 195 grams of water at 27 deg C absorbs 289 kJ of heat from a mountain climber's stove at an elevation where the boiling point of water is 91 deg C, is the amount of energy sufficient to heat the water to it's boiling point? And how many kJ are required to boil the water?

Question

anonymous · Answer

J=CpxGxC
Energy=Specific heat times grams times celcius degrees.

anonymous · Answer

91-27 = 64 celcius degrees times 195 times specific heat of water which I believe is 4.184? so just multiply all of them and you shall get your answer! :)