given m = 0.305
temperatures are 62.4 , -20.4 , 27
the mass of the ice is say M
Qheat energy= m X C X dT
Qliquid= 0.305 X 4190 X (62.4 - 27 )
= 45239.43 J
the heat gained by ice is
Qice + Qfusion + Qliquid
Qice = M X 2100 X ( 0 - ( -20.4 ) )
= 42840 M
Qfusion = M X 3.34 X 105
Qliquid = M X 4190 X 27
= 113130 M
loss of heat = gain in heat
Qliquid = Qice + Qfusion + Qliquid
45239.43 = 42840 M + M X 3.34 X 105 + 113130 M
45239.43 = 489970 M
M = 45239.43 / 489970
M = 0.09233 Kg
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