Chemistry

If 118 g of ice at 0.0°C is added to 1.55 L of water at 88°C, what is the final temperature of the mixture?

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asked by Sam
  1. heat absorbed by melting ice + heat to raise water from zero C + heat lost by 88 C water.= 0
    [mass ice x heat fusion] + [mass melted ice x specific heat water x (Tfinal-Tinitial)] + [mass separate water x specific heat water x (Tfinal-Tinitial)] = 0.
    Substitute the numbers and solve for Tfinal.

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  2. how do i get the mass of separate water, if it is in L?

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    posted by Sam
  3. 1.55L is 1550 mL (1550 cc) and if the density of water is 1.00 g/mL, then the mass of the water is 1550 grams. Most problems of this nature usually state that the water is to be considered as having a density of 1 g/cc.

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  4. um...one more question, what do you mean by "mass of melted ice"? how is it determined?

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    posted by Sam
  5. Let's see now. If I had 1 g ice and it melted into water, it would form 1 g water. So grams ice and grams melted ice must be the same. right?

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