Posted by chuck on Wednesday, September 12, 2012 at 9:34am.
I think you have two problems here. The first one ends with "was there?" Then the second one begins. The first one must take into account the fact that the 150 g ice melts at zero C and that water must be raised to 14 C while the other water is being reduced by 1 C (from 15 C to 14 C). The second problem ignores the 150 g H2O produced by the melted ice.
Are you working in J or calories? Calories?
150 g x heat fusion (80 cal/g) = 12,000 calories.
If you want the water to go from 15 C to 14 C, then you can remove 12,000 calories. Heat capacity of H2O is 1 cal/g*C, that means 12,000 g H2O. That ignores the 150 g H2O at zero that comes from the melted ice and that must be raised to 14 C. That could remove another 150 x 14 = 2100.
The two answers are 12,000 g H2O ignoring the melted ice or 14,100 g H2O including the ice.
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