A 95 gram piece of metal is heated to 98°C and placed into 200mL of water insulated at 20°C. If the equilibrium temperature is 27.20, find:

Question

The heat gained by the water. calories
The specific heat of the metal. cal/(g°C)

Hint: 1 mL water has a mass of 1 gram

I have no idea how to begin any of this, much less how to determine the answers for the two parts. Can someone help please?

Answer 1

heat gained by water:
masswater*specificheat*deltatemp
200g*4.18j/g*(27.2-20)=...

heat gained by water+heat change by metal=0
above heat calculation+95g*cmetal*(27.2-98)=0
solve for cmetal

Answer 2

6,019.2 calories for the heat gained by water is incorrect. That has been my problem all along. Can't figure out the correct answer.
Thanks, though.