# HW11S - Owlia Arria Homework 11 Due midnight Inst...

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Owlia, Arria – Homework 11 – Due: Nov 27 2007, midnight – Inst: Vandenbout 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Use thermodynamic values given in the problems when they are given. If they are not given, you need to use your textbook and look up the needed values. Please use the following Water Data C s , ice = 2 . 09 J/g C C s , water = 4 . 184 J/g C C s , steam = 2 . 03 J/g C Δ H fusion = 334 J/g Δ H vapor = 2260 J/g 001 (part 1 of 1) 10 points 70 g of ice and 70 g of water are at equil- brium in an insulated container. 47 kJ of heat are delivered to this mixture via an electric heater. What is the final temperature of wa- ter in the container? Correct answer: 40 . 3 C. Explanation: Calculate the amount of heat required to melt all of the ice. Subtract that from the 47 kJ delivered. The remaining Joules will heat the now 140 g of water. 002 (part 1 of 1) 10 points Consider an insulated system containing 100 g of liquid water and 100 g of ice at equi- librium under atmospheric pressure. A total of 12 g of steam at +110 C is admitted to the system. What is the phase composition and temperature when equilibrium is reestab- lished? 1. 209.0 g of water and 3.0 g of ice at 0.0 C correct 2. 208 g of water and 4 g of ice at 0.0 C 3. 212.0 g of water at 1.3 C 4. 212.0 g of water at 1.6 C 5. 212.0 g of water at 38.7 C Explanation: m water = 100 g m ice = 100 g m steam = 12 g T = 110 C For these types of problems, it is reasonable to assume that the final temperature must be somewhere between 0 C and 110 C. Given the relatively large amount of ice compared to the steam, the final temperature is likely to be closer to 0 C. So, the probable final scenarios are these three: (1) The heat released when the steam is cooled is just enough to melt all the ice. Thus at the end, 212 g of water (100 g ini- tially water, 100 g of ice that has melted, and 12 g of steam which has condensed to water) will remain at 0 C; (2) The heat released when the steam is cooled is greater then the amount needed to melt all the ice. The “excess” heat then warms the 212 g of water to some temperature above 0 C; (3) The heat released when the steam is cooled is insufficient to melt all the ice. Thus, an ice/water mixture will remain at 0 C. Scenario (1) is unlikely, but it does repre- sent a convenient situation to consider. Our approach to this problem then is to calculate the amount of heat required to melt all the ice, and also calculate the amount of heat re- leased when the steam is cooled to water at 0 C. Whichever value is larger will let us know whether scenario (2) or (3) has taken place, and we will be able to identify what the final conditions are. Ice will melt at a constant temperature of 0 C, governed by the following equation: q = Δ H fus (m) where m = mass or moles, so q = (334 J / g)(100 g) = 33400 J .

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Owlia, Arria – Homework 11 – Due: Nov 27 2007, midnight – Inst: Vandenbout 2 Thus 33400 J would be required to melt all the ice. Now let’s consider how much heat is
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