Question

A piece of metal (75 g at 98.0°C) was dropped in water (95 g at 22.0°C) and the final temperature was measured to be 31.2°C. If the specific heat of water is 4.18 J/g°C. What is the heat capacity of the metal?

Answer 1

heat lost by metal + heat gained by water = 0
[mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial) = 0

Solve for specific heat metal

Note that you have calculated the specific heat capacity of the metal. The problems asks for heat capacity so you must convert specific heat capacity (that's J/gram*C) to heat capacity (that J/C)