Answer :

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

Mass of ammonia be= M

Mass of water =2000g

Temperature change, ΔT=20^{o}C-0^{o}C=20^{o}C

Amount of heat energy released in colling 2 Kg water from 20^{o}C to 0^{o}C

Q_{W}=MCΔT

Where

M is the mass of water

C is specific at constant volume

ΔT is the temperature difference

Putting the values in the above formula, we get

Q_{w}=2000×1×20=40000 cal

^{o}C to ice at 0

^{o}C

Q

_{i}=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

Q

_{i}=2000×80=160000 cal

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

Q = m× L=Q_{i} +Q_{w}

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

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