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

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:
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?

bobpursley · Answer

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

Anna · Answer

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.