Jawabanmu

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2014-08-31T23:13:16+07:00
ΔTf = kf . m . i
     = 1,86 .  \frac{1,74}{174} .  \frac{1000}{50} . 3
     = 1,86 . 0,01 . 20
     = 0,372°C

Tb = ΔTb + Tb°
     = 0,372 + 100
     = 100,372°C
seharusnya Tf = 0 - deltaTf = 0 - 0,372 = -0,372