Chemistry problem please help!

Question

If a cup of tea (190.0 ml, 86.0 celcius) is cooled by adding an ice cube (30.0 g, -10.0 celcius), what is the final temperature of the tea? The heat required to melt ice (heat of fusion) is 80.0 cal/g

Answer 1

Okay, so first we'll set this up by common sense.  We're going to assume that the solids in the tea have no effect on the heat, so 1 calorie will still raise/lower 1 ml of water (or tea) by 1 degree celcius.  We're also, for convenience, going to remember that 1 g of ice melts to 1 g of water which has a volume of 1 ml.  Ready to set this up?

 

We know the ice will ALL melt, so that's going to suck up 80*30 calories.  We also have 30 grams of soon-to-be-unfrozen water at -10 degrees, for another 10*30 calories gone to take that water up to zero.  Other than that, we have 190 ml at 86 degrees, for 190*86 = 16340 (positive) heat calories above zero celcius in the mix. 

 

16340 -2400 -300 = 13640 calories above zero celcius left at equilibrium, in a cup containing 190+30 = 220 ml of tea.

 

13640/220= 62 degrees celcius.