Question

A 3.53g sample of (NH4NO3) was added to 80.0mlof water in a constant pressure calorimeter of negligible heat capacity. As a result the temperature of the water decreased from 21.6C to 18.1C Calculate the heat of solution of NH4NO3. Help me ....

Answer 1

Mass of water = 80 x 18
= 1440 grams

Specific heat of water = 4.186 J/gK

Heat absorbed from water = mCpΔT
= 1440 x 4.186 x (21.6 - 18.1)
= 21.1 kJ

Moles of NH₄NO₃ = 3.53 / (14 + 4 + 14 + 16 x 3)
= 0.044 mol

Heat of solution = 21.1/0.044 kJ/mol
= 480 kJ/mol