A 25 g glass tumbler contains 450 mL of water at 24掳C. If four 11 g ice cubes each at a temperature of -3掳C are dropped into the tumbler, what is the final temperature of the drink? Neglect thermal conduction between the tumbler and the room.|||So you have 44 grams of ice at 0.5 calories per degree = 44 times 0.5 times 3 = 66 calories to warm the ice to the melting point plus 44 grams times 80 (The heat of fusion of ice) = 3520 + 66 = 3586 calories = water at zero c If the water cools 10 degrees c ; that is 4500 plus 50 calories to cool the tumbler 10 degrees c = 4550 c(it looks like we are close with the 14 degree estimate) We need to warm the 44 grams of ice water by 14 degrees c = 616 calories + 3586 = 4202 calories. The answer is likely about 15 degrees c. You can repeat my calculation for 15 degrees instead of 14, the do a third estimate for perhaps 15.1 depending on the discrepancy at 15 degrees. Alternately you can likely find the formulas in your text book, and write an equation making them equal for an algebraic solution. You can alternately plot a graph of the two curves and see where they intersect near 15 degrees. I guessed the specific heat of the tumbler is 0.2 calories per gram. Likely it is somewhat more. There are many kinds of glass, so silica plus calcium oxide plus sodium oxide is likely the kind the tumbler is made of. Possibly called common glass. Good luck if you need 4 significant figures. Neil

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