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Introductory Physics Homework Help
Temperature after mixing ice and water
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[QUOTE="takando12, post: 5470584, member: 557635"] I understand now. So I divide into two stages and then check. 1) Raise temp of ice to melting point Q=CmΔT=0.5*20*5 =50J 2) Melt the ice Q=Lm =80*5 =400J Total heat required = 450J 3) Check if water can provide this heat Q=CmΔT= 1*5*30 =150J. It cannot.So only some of the ice will melt? m=Q/L =100/80=1.25 g of ice melts. And the temp sticks to 0°C. If I was given say 6g of water instead, I would be able to release 180J. So that means 30 J will go into further raising the temp of the melted ice form 0 to 3°C? (that is adding the total mass to be 10g of water) .Is that correct? [/QUOTE]
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Introductory Physics Homework Help
Temperature after mixing ice and water
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