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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) |
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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 . |
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