Respuesta :

Answer:

the heat change to 10kg of ice to water 0

so quantity of heat required is answer: 5460 J.

hope its helps!

Answer:

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