Posted by Lauren on Monday, November 5, 2012 at 3:53pm.
a. q1 needed to melt all of the ice
q1 = mass ice x heat fusion.
q2 = mass H2O x specific heat H2O x delta T. (and the specific heat H2O is NOT 1.184).(Consider delta T = 80 which is the maximum heat that can be released). There is more than enough to melt all of the ice. This work answers a and the conclusion answers c.
For b, just put all of this together.
q1 = heat to melt ice.
q1 = [mass ice x heat fusion]
q2 = heat to raise temperature of melted ice to final T.
q2 = mass melted ice x specific heat H2O x (Tf-Ti) (of course Ti is zero).
q3 = heat released by the 80 C water.
q3 = [mass H2O x specific heat H2O x (Tf-Ti) (Ti of course is 80.)
Now just add all of these together. The sum is zero.
q1 + q2 + q3 = 0
Substitute q1, q2 and q3 from above and solve for Tf. I estimated the value and obtain something around 20 C or so.
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