Specific heat of water is 4.179 J/g˚C (http://www.iun.edu/~cpanhd/C101webnotes/matter-and-energy/specificheat.html)
Thus Gooperman! has a specific heat of 20895 J/g˚C
The formula that dictates the change in temp is ∆Q=mc∆T
where m is mass
Q is heat
c is specific heat
And T is temp in celcsisu
Gooperman melts at 1500˚K or 1226.85˚C
Assuming he starts at room temperature is at most 25˚C
So ∆T is 1201.85
Thus it can be modeled as ∆Q=2.563503075x10^7(m)