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=10×0,5×40
=200 J
Q=m.l
=10×80
=800 J
Q=m.c.∆t
=10×1×100
=1000 J
*kalor yg diperlukan=200+800+1000=2000 J
=10×0,5×40
=200 J
Q2 =m.L
=10×80
=800 J
Q3=m.c.∆t
=10×1×100
=1000 J
kalor total = Q1 + Q2 + Q3
kalor total =200+800+1000=2000 J