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.

 

[DeeNos]

Your approach and calculations seem correct. The final temperature of 49.45 Celsius does seem unexpected for a cold pack. One thing to consider is that the heat of dissolution of NH4NO3 is not the only factor affecting the temperature change in the pack. Other factors such as the heat capacity of the pouch and the rate of heat transfer between the pack and the environment may also play a role. It is also possible that there may be some experimental error in the temperature measurements. I would suggest double checking your calculations and also considering these other factors when trying to explain the unexpected result.