Question

Liquid ammonia (boiling point = -33.4 degrees Celsius) can be usedas a refrigerant and heat transfer fluid. how much energy is neededto heat 25.0 g of NH₃ (liquid) from -65.0 degreesCelsius to -12.0 degrees Celsius ?

specific heat capacity, NH3 (liquid)   4.7 J/ (g x k)
specific heat capacity, NH3 (gas)      2.2J/ (g x k)
heat ofvaporization                  23.5 kJ/mol
molarmass,                    17.0 g/mol

correct answer: 39 kJ

please show me step by step how to solve this problem, i'm notgetting the correct answer when i try to solve it. thanks a bunch:)!

Answer 1

      step 1 heating liquidammonia from -65^oC to -33.4^oC

           q1 = m s ΔT

                 = 25.0 g * 4.7 J/g.K * -33.4^oC - (-65^oC)

                = 3713 J

    step 2: evaporating 25 g ofNH₃ at -33^oC

         q2 = (25 / 17)mol * ΔH_vap

               = 1.470 mol * 23.5 x 10³ J /mol

               = 34558.8 J

  

   step 3 : heating NH3(g) from-33.4^oC to -12.0^oC

        

          q3 = m s ΔT

                =25.0 g * 2.2 J /g.K * -12.0^oC -(-33.4^oC)

               = 1177 J

     ∴ total heat energy = q1 + q2 +q3

                                    = 39448.8 J

                                     =39.4 kJ