A beaker contains 580 g of ice.into the beaker is passed steam till ice left is 500g.The water collected is 90g.If the specific heat of ice is 80 cal /g.
Find L of vaporisation of steam

Asked by lovemaan5500 | 13th Nov, 2017, 06:17: PM

Expert Answer:

 

Heat gain by 80 gm of ice - from 0°C ice to  0°C water = 80 x 80 = 6400 cals/gm

Heat lost by 10gm steam – from 100°C steam to 0°C water = 10L+10(100-0)

L – latent heat of condensation of steam

Heat loss by steam = heat gain by ice

10L+1000 = 6400.

So we get L = 540 cals/gm = 545 x 4.2 = 2268 kj/kg


 


Answered by Gajendra | 6th Dec, 2017, 05:13: PM

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