m1 = 150 g , m2 = 300 g,
Specific heat of water Cp =4.186 J/g/K
Latent heat of ice L = 334 J/g
Heat energy gain by ice = Heat eenrgy loss by water
m1L + m1Cp(DT) = m2Cp(DT)
(150*334) +(150*4.186*(T-0)) = - (300*4.186*(T - 50))
final temperature T =6.734 oC
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