Calculate ΔHrxn for Liquid -> Solid Water at -5°C

[Neophyte]

Homework Statement

What is the heat change in J associated with 80.7 g of liquid water at 5.00 ° C changing to solid water at -5.00 °C?

Homework Equations

c(H2 O) (liq) = 4.184 J/(g.K)

c(H2O) (s) = 2.09 J/ (g.K)

DHfus(H2 O) = 6.02 kJ/mol

The Attempt at a Solution

Δt = 5°

While liquid
(80.7g)(4.184J/g.k)(5°) = 1688.2 J
(-6.02kJ/mol)(80.7/18 = 4.48 mol) = -27.0 kJ
(80.7g)(2.09J/g.k)(5°)= 843.3 J

2531.5 - 27000
-2.44E4
(correct answer is -2.95E-4)

I mean I was pretty sure that is how you went about doing this type of problem :/.
The Δt = 5° came from (5 - 0) (and 0-(-5))

 

[chemisttree]

I believe you are messing up in your sign convention. You are removing heat throughout this process and thus all of your signs must be negative. Delta T is -5, not 5.