Question #1645

How much ice at -10C must be added to 0.50 Kg of water at 20 C if it all to melt leaving only water at 0 C?

Expert's answer

Let's construct heat ballance equation:

C_{ice}*m*(T_{1}-T_{0}) + C_{fnice} * m = C_{water}*m_{water}*(T_{2}-T_{0})

where C_{ice} is a specific heat of ice, 2.108 J/gK , C_{water} is a specific heat of water of 4.187 J/gK, C_{fnice} - the heat of ice fusion 333.55 J/g, T_{0} = 0C (melting point of ice), T_{1} = -10C, T_{2} = 20C, m_{water} = 0.5kg=500g.

21.08m + 333.55 m = 41870

**m = 118.1 g = 0.118 kg of ice.**

C

where C

21.08m + 333.55 m = 41870

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