Question

Liquid ammonia is used in ice factory for making ice from water . If water at 20°c is to be converted into 2kg ice at 0°, how many grams of ammonia are to be evaporated ?(given the latent heat of vaporisation of ammonia is 341 cal/g)

Answer 1

Latent heat of fusion of water = 334 Joules / gram = Lf

Heat capacity of water = 4.186 Joules / gram

Latent heat of vaporization of Ammonia in Joules per gram : Lv

1 calories = 4.184 joules

341 × 4.184 = 1426.74 Joules / gram

Heat lost = heat gained

Mlv = mcФ + mlf

Mass of Ammonia = M

Mass of water = 2000g

Δ Temperature = Ф = 20°

1426.74M = 2000 × 20 × 4.186 + 2000 × 334

1426.74M = 835440

M = 835440/1426.74

M = 585.56 g

Answer 2

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