Answer:
the heat change to 10kg of ice to water 0
so quantity of heat required is answer: 5460 J.
hope its helps!
Q=ml+mc∆+ml'
=10*80+10*1*(100-0)+10*540
=800+1000+5400
=7200cal.7.2kcel
L=heat of fusion of ice
L'=heat of vapourisation of water
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