Can someone explain in detail all the steps for thisproblem? A heat of solution

Question

Can someone explain in detail all the steps for thisproblem? A heat of solution of ammonium nitrate is 26.2 kJ/mol. If a5.368g sample of NH4NO3 is added to40.0mL of water in a calorimeter at 23.5oC, what is theminimum temperature reached by the solution? (specific heat of water = 4.18 J/goC             heat capacity of the calorimeter = 650. J/oC) Can someone explain in detail all the steps for thisproblem? A heat of solution of ammonium nitrate is 26.2 kJ/mol. If a5.368g sample of NH4NO3 is added to40.0mL of water in a calorimeter at 23.5oC, what is theminimum temperature reached by the solution? (specific heat of water = 4.18 J/goC             heat capacity of the calorimeter = 650. J/oC)

Explanation / Answer

1st heat produced: 5.368 g / (80 g/mol) = .0671 mol (26.2 Kj/mol)(.0671 mol) = 1.758 KJ = 1758 J (4.184 J/gC)(40 g)(t) + (650 J/C)(t) = 1758 J t = 2.15 Celsius Finally, 23.5 - 2.15 = 21.35 Celsius

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