Liquid ammonia (boiling point = -33.4 degrees Celsius) can be usedas a refrigera

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:)!

Explanation / Answer

      step 1 heating liquidammonia from -65oC to -33.4oC            q1 = m s T                  = 25.0 g * 4.7 J/g.K * -33.4oC - (-65oC)                 = 3713 J     step 2: evaporating 25 g ofNH3 at -33oC          q2 = (25 / 17)mol * Hvap                = 1.470 mol * 23.5 x 103 J /mol                = 34558.8 J       step 3 : heating NH3(g) from-33.4oC to -12.0oC                    q3 = m s T                 =25.0 g * 2.2 J /g.K * -12.0oC -(-33.4oC)                = 1177 J      total heat energy = q1 + q2 +q3                                     = 39448.8 J                                      =39.4 kJ

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