Mass of amonia = 34.06 gm
Molar mass = 17.03 gm
Number of moles, n = mass/ molar mass = 34.06/17.03 = 2
mol
Step 1: Raise temperature of liquid amonia from -40 oC
to -33.4 oC
Q1= n*C*(Tf - Ti)
   = 2*80.8*(-33.4 - (-40))
   =1066.56 J
Step 2: Vapourise the liquid ammonia at -33.4
oC
Q2= n*C
      = 2*23.35 KJ
      = 76.7 KJ
      = 76700 J
Step 3: Raise temperature of vapour amonia from -33.4 oC
to 25 oC
Q3= n*C*(Tf - Ti)
   = 2*35.06*(25-(-33.4))
   =4095.01 J
Total heat absorbed = Q1+Q2+Q3
                                        Â
= 1066.56 + 76700 + 4095.01
                                        Â
=81861.57 J
                                         Â
= 81.86 KJ
Answer:Â Â 81.86 KJ
