The formula is #Q = m*c*DeltaT#. Variables: #m " is mass, "# #DeltaT " is temperature change"#. The #c# is called specific heat. It is a characteristic of the material -- water in this case. You find it in a table, or your book, or etc.
For water, hyperphysics gives it as #4.186 J/(gm*K)#. For this problem which gives the mass in kg, it would be simpler to use #4.186 (kJ)/(kg*K)#. But, I have one more simplification: because #"a change of " a^@ K = "a change of "a^@ C#, it is valid, and even simpler, to use #4.186 (kJ)/(kg*C)#.
Plugging the data into
#Q = m*c*DeltaT#
#Q = 2 cancel(kg) * 4.186 (kJ)/(cancel(kg) * C)*(100^@ C - 15 ^@ C)#
#Q = 2 * 4.186 (kJ)/cancel(C)*85cancel(C) = 712 kJ #
I hope this helps,
Steve
