Can someone kindly double check my heat and energy problem?

[hockeyfan]

Homework Statement

Instant cold packs used to treat athletic injuries contain solid NH4NO3 and a pouch of water. When the pack is squeezed, the pouch breaks and the solid dissolves, lowering the temperature because of the endothermic reaction

NH4NO3(s) + H2O(l) ---> NH4NO3(aq) dH = 25.7 kJ

What is the final temperature in a squeezed cold pack that contains 50.0g of NH4NO3 dissolved in 125ml of water? Assume a specific heat of 4.18J/g*C for the solution, an initial temperature of 27.5 Celsius, and no heat transfer between the cold pack and the environment.

Homework Equations

Equation I used: q = m x c x dT

The Attempt at a Solution

My steps:
1) 50.0g NH4NO3 / 80.02g NH4NO3 = 0.625 moles NH4NO3
2) 0.625 moles NH4NO3 x 25.7 kJ = 16.1
3) 16.1 x 1000 = 16058 for q
4) 125ml --> 125g via density. 125g + 50.0g = 175 total g.
4) Plug in numbers into equation: 16058 = 175 x 4.18 x (Tf - 27.5) *Unsure about this part
5) 16058 = 731.5 Tf - 20116.25
6) 16058 + 20116.25 = 731.5 Tf
7) 36174.25 / 731.5 = Tf
8) 49.45 Celsius. So... I'm pretty certain cold packs aren't suppose to get hot, or they would be called hot packs (haha). On a more serious note I really want to figure this out. Help please!

 

[Borek]

Positive ΔH means reaction takes energy from the outside, not gives it away.

In other words: sign convention.

 

[hockeyfan]

So, I change 16058 to -16058, follow the same process, and I receive an answer that's approximately 5.5. Is this correct?

 

[Borek]

At least it SOUNDS correct.