1. The problem statement, all variables and given/known data 2.0kg of a metal (for which c = 620 J kg K ) is immersed in 1.5kg of water, initially at 20◦ . What is the final temperature if 8.0 × 104J are added as heat? 2. Relevant equations Q=mcΔT 3. The attempt at a solution So I was able to get the correct answer what I did was: Qadded heat+mcΔTiron+mcΔTwater=0 -8×104 +2(620)(Tf-293)+1.5(4200)(Tf-293)=0 doing the algebra will give you 303.6 K which is correct...but I was just wondering why is the heat added negative?