1.

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? ( Given: The latent heat of vapourization of ammonia = 341 cal/ g)

Answer»

Solution :Given:Latent heat of melting of ice =`L_"melt"`=80 cal/G
Latent heat of VAPOURIZATION of ammonia = 341 cal/g
Temperature of liquid water =`20^@C`
Temperature of ice = `T_"ice"=0^@C`
Solution:Heat RELEASED during conversion of liquid water into water at `20^@ C , =m_(H_2O) xx L_"melt"` = 2 x 100 x 80 ....(i)
Heat released during conversion of water of `20^@`C into ice at `0^@C`
`=m_(H_2O) xx DeltaT xxC`
= 2 x 1000 x (20 - 0) x 1
= 2 x 1000 x 20 ...(ii)
Total heat gained by ice,
=(2 x 1000 x 80 ) + ( 2 x 1000 x 20 ) [ from (i) and (ii) ]
=2,00,000 cal
AMOUNT of ammonia required to be evaporated
`m_"Ammonia " xx L_"vap of ammonia " `=2,00,000
`m_"Ammonia"=200000/341`=586.4g
586.4 g of ammonia are to be vaporated .


Discussion

No Comment Found