Posted by James on Wednesday, February 10, 2010 at 6:03pm.
Calculate the heat that must be removed from the liquid H2O to bring it to 0 C and freeze it.
That would be 125*80 + 125*10*1
= 11,250 Cal
Then calculate the heat that must be added to the ice to bring it to 0 C without melting it. That would be
35*0.5 * 500 = 8750 Cal
The ice cannot absorb enough heat from the water to freeze it all. It would take another 2500 Cal of heat removal to freeze all the water. That means that 2500/80 = 50 ml (or 50 g) of the water will remain liquid at 0 C. the other 75 g will be added to the 500 g of ice.
You end up with 575 g of ice and 50 ml of water at 0 C.
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