Question

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

Answer 1

Hello Dear.

Here is the answer⇒

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Mass of Ice = 2 kg.
Thus, Mass of water at 20°C  = 2 kg.

Specific Heat Capacity of Water = 4200 J/kg°C

Change in temperature(Δt) = (20 - 0)
                                              = 20 °C

Thus, Amount of Heat Energy Required to convert water at 20° to 0° C
 

= mass × Specific Heat Capacity × Change in temperature
= 20 × 4200 × 20
= 400 × 4200
= 168 × 10⁴ J.

Latent Heat of Fusion of Ice = 336000 J/kg.

Amount of Heat Energy Required to Convert water into ice at 0°
= mass × Latent Heat of Fusion
= 20 × 336000
= 672 × 10⁴ Joule

Thus, Total Heat Energy = 168 × 10⁴ + 672 × 10⁴
                                       = 10⁴(168 + 672)
                                       = 10⁴ × 800
                                       = 8 × 10⁶ J.


Let the Mass of the Ammonia be m kg.

We know the Latent heat of Vaporization of Ammonia = 1400 × 10³ J/kg.


By the Principle of Calorimetry,

Heat Given By m kg of ammonia in Vaporization = Heat Taken by the Water to change into Ice.

       ⇒ m × 1400 × 10³ = 8 × 10⁶
       ⇒ m = (8/1400) × 10³
       ⇒ m = 5714.3 g.


 Thus, the mass of the ammonia required is 5714.3 grams.


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Hope it helps.