1. The problem statement, all variables and given/known data What is the final temperature in a squeezed hot pack that contains 23.9 g of LiCl dissolved in 107 mL of water? Assume a specific heat of 4.18 J/(g⋅∘C) for the solution, an initial temperature of 25.0 ∘C, and no heat transfer between the hot pack and the environment. 2. Relevant equations q=mcΔT LiCl(s)⟶Li+(aq)+Cl−(aq)ΔH=−36.9kJ 3. The attempt at a solution 1. Change 23.9 g LiCl to mol. I get .564 mol LiCl 2. Use .564 mol LiCl x -36.9kJ to get -20.8 kJ for this reaction 3. Add masses of water and LiCl to get 130.9 g total. 4. Change -20.8 kJ to J: -20800 J 5. use q=mcΔT to get -20800J=(130.9g)(4.18g⋅°C)(Tf-25.0°C) 6. Find Tf to be -13.0°C. 7. Answer is wrong and I am confused. Please could someone help out?