Grade Calorimetry II Name Date Data: water=4190ice = 2100ater = 334 x 10%/ A water cooler has m = 12.0 kg of water at a

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Grade Calorimetry Ii Name Date Data Water 4190ice 2100ater 334 X 10 A Water Cooler Has M 12 0 Kg Of Water At A 1 (28.99 KiB) Viewed 53 times

Grade Calorimetry II Name Date Data: water=4190ice = 2100ater = 334 x 10%/ A water cooler has m = 12.0 kg of water at a temperature T = 54°C. A mass m = 2.10 kg of ice at temperature T₁ = -15°C is put into the water. After all the ice melts determine the final equilibrium temperature T, by going through the following steps: a) Write a term for the heat gain in warming the ice from 7, to its melting point. Use symbols not numbers. No numbers are allowed until the last step. b) Write a term for the heat gained in melting the ice. c) Write a term for the heat gain in warming the melted ice from its melting point to T d) Write a term for the heat gain from cooling the water from its initial temperature to T e) Put all of these together to construct a single equation for the conservation of heat energy. f) Find an algebraic solution for the final temperature T g) Plug numbers with units into this equation to find the final temperature in °C. Note: you should do the entire calculation in one step, with NO intermediate results. If you can't, practice until you can. h) Finally, is this a reasonable answer? Briefly, why or why not?