Heat and temperature question with calorimetry

[Lisa Marie]

Homework Statement

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?

Homework Equations

Q=mcΔT

The Attempt at a Solution

So I was able to get the correct answer what I did was:
Q_{added heat}+mcΔT_iron+mcΔT_water=0
-8×10⁴ +2(620)(T_f-293)+1.5(4200)(T_f-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?

 

[rude man]

Homework Statement

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?

Homework Equations

Q=mcΔT

The Attempt at a Solution

So I was able to get the correct answer what I did was:
Q_{added heat}+mcΔT_iron+mcΔT_water=0
-8×10⁴ +2(620)(T_f-293)+1.5(4200)(T_f-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?

I assume the metal initial temp. was 20C also.
Your equation is conceptually somewhat unwieldy. More obviously,
heat gained by water + heat gained by metal = external heat added
which is your equation also if you move the external heat to the rhs of your equation.