A calorimeter consists of a massless beaker that contains 20.0 g of solid water

Question

A calorimeter consists of a massless beaker that contains 20.0 g of solid water (i.e., ice) at -30 degrees Celsius . Then 100.0 g of iron at 70.0 degrees celsius are added to the calorimeter. a) Find the final temperature of the calorimeter after equilibrium is reached. If the final temperature is 0 degrees Celsius, what fraction of the ice is melted? b) During the process of reaching thermal equilibrium, how much heat is absorbed by the H2O? by the iron? What is the sum of these two heats? Why must the sum have this value?

Explanation / Answer

To solve this we need to have heat gained by the ice equal heat lost by the iron. Ice takes less heat (per gram and per degree) to warm than water. Ice at zero C takes a lot of heat the melt. Water takes about twice as much heat to warm than ice. Iron takes less less heat to warm than water. Let's get some values Heating ice 2.11 J/gK Melting ice at 0C 334 J/g Heating water 4.18 J/gK Heating iron 0.450 J/gK Iron can give 70K*100g*0.450J/gK = 3150 J Ice needs 20g*30K*2.11J/gK = 1266 J to get to zero, so the iron will get all the ice to 0C since it can give 1266J to the water. When it does, there are (3150-1266)J left to melt ice. It would take 334J/g*20g = 6,680 J to melt all of the ice and we only have (3150-1266)J, so the final temperature will be zero, and there will be ice left. (3150-1266)J/6,680 J = 0.282 or 28% of the ice is melted. Iron lost = mass*heat capacity*change in temperature = 100g*0.450J/gC*(70-0)C The water absorbed all of the heat that the iron lost. The iron lost all of the heat that the water absorbed. The sum is zero since the calorimeter didn't exchange any heat with the room.

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