Can someone kindly double check my heat and energy problem?

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Discussion Overview

The discussion revolves around a homework problem involving the calculation of the final temperature of a cold pack containing solid NH4NO3 and water, focusing on the endothermic dissolution process and the associated heat transfer calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • The initial setup of the problem includes the dissolution of NH4NO3 in water, which is described as an endothermic reaction with a given enthalpy change.
  • One participant outlines their calculation steps, including converting grams of NH4NO3 to moles and calculating the heat absorbed by the solution.
  • Another participant points out the sign convention for ΔH, indicating that a positive ΔH means the reaction absorbs energy from the surroundings.
  • A subsequent participant modifies the heat value to a negative sign and recalculates, arriving at a different final temperature of approximately 5.5 degrees Celsius.
  • Another participant expresses uncertainty about the correctness of the modified calculation.

Areas of Agreement / Disagreement

Participants express differing views on the sign of the heat transfer and its implications for the final temperature calculation. There is no consensus on the correct final temperature, as calculations yield different results.

Contextual Notes

Participants have not resolved the implications of the sign convention on the calculations, and there are unresolved steps in the mathematical reasoning presented.

Who May Find This Useful

Students working on thermodynamics problems, particularly those involving endothermic reactions and heat transfer calculations in chemistry or physics contexts.

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

In other words: sign convention.
 
So, I change 16058 to -16058, follow the same process, and I receive an answer that's approximately 5.5. Is this correct?
 
At least it SOUNDS correct.
 

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