The specific heat of a certain type of metal is 0.128j/(g. c). what is the final temperature if 305.j of heat is added to 72.7g of the metal initially at 20.0°c?

E answer is -60.57 = -60.6 KJ.
CaC2(s) + 2 H2O(l) ---> Ca(OH)2(s) +C2H2(g) H= -127.2 KJ
Hf C2H2 = 226.77
Hf Ca(OH)2 = -986.2
Hf H2O = -285.83
Now,

add them up. 226.77 - 986.2 + (2*285.83) = -187.77 
Add back the total enthalpy that is given in the question
 -187.77+127.2 = -60.57