Question 15.

**Solve the following problems:**

**Liquid ammonia is used in an ice factory for making ice from water. If the water at 20oC is to be converted into 2 kg ice at 0oC, how many grams of ammonia is to be evaporated?(Given: The latent heat of vaporization of ammonia= 341 cal/g)**

Answer:

Given: Latent heat of vapourisation of Ammonia =341 cal/g

Mass of ammonia be= M

Mass of water =2000g

Temperature change, Î”T=20oC-0oC=20oC

latent heat is the heat required by the ammonia for the liquid ammonia to ammonia vapours.

Amount of heat energy released in colling 2 Kg water from 20oC to 0oC

QW=MCÎ”T

Where

M is the mass of water

C is specific at constant volume

Î”T is the temperature difference

Putting the values in the above formula, we get

Qw=2000×1×20=40000 cal

Amount of heat energy released during the conversion of 2kg of water at 0oC to ice at 0oC

Qi =M×L

Where

M is the mass of the water

L is the latent heat of conversion of water to ice

Putting the values in the above equation, we get

Qi=2000×80=160000 cal

The heat gained by ice due presence of ammonia will be:

Q = m× L=Qi +Qw

Where

m is the mass of the ammonia

L is the latent heat of the ammonia

On solving, we get

m× L=160000+40000=200000 cal

Thus, the process of conversion of 2kg of water to ice will require evaporation of 586.5 g of ammonia.