if you have 125.0 g of water at 100.0 degrees celsius, how much energy must be added to vaporize all of it

Answer 1

Since the initial temperature of the water is already at 100ºC, the first energy used will be for the phase change from liquid to gas. For this you need to know the ∆Hvaporization. Looking it up, you find it to be 2260 J/g (you may find a different value depending on the source). Then you can use q = m∆Tvap

q = (125.0 g)(2260 J/g) = 282,500 J = 282.5 kJ