First, determine q for the reaction. That is
25700 J x (45.5/molar mass NH4NO3) = approx 15,000 J.
Then
q = [mass H2O x specific heat H2O x (Tfinal-Tinitial)
q from above
mass H2O = 125g
specific heat H2O given
Solve for Tf
Ti is given
NH4NO3(s)+H2O(l)→NH4NO3(aq) ΔH = +25.7 kJ.
What is the final temperature in a squeezed cold pack that contains 45.5g of NH4NO3 dissolved in 125 mL of water? Assume a specific heat of 4.18J/(g⋅∘C) for the solution, an initial temperature of 29.0∘C, and no heat transfer between the cold pack and the environment
1 answer
1 answer