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Worksheet 02 Key

# Worksheet 02 Key - 1 Spring 2007 CH 307 Worksheet 2 100 g...

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Spring 2007 CH 307 Worksheet 2 1. 100 g of ice at -25°C is heated to steam at 125°C. For water, the specific heats are c ice = 2.093 J/g°C, c water = 4.186 J/g°C, and c steam = 2.009 J/g°C. The enthalpy changes are H fusion = -335.5 J/g and H vaporization = 2.26 kJ/g. What is H sys for this process? Answer: Divide the process into several steps: Heat ice to 0°C: H = mc T = (100 g)(2.093 J/g°C)(25°C) = 5.232 kJ Melt ice: H = -m H fusion = -(100 g)(-335.5 J/g) = 33.55 kJ Heat water to 100°C: H = mc T = (100 g)(4.186 J/g°C)(100°C) = 41.86 kJ Boil water: H = m H vaporization = (100 g)(2.26 kJ/g) = 226 kJ Heat stem to 125°C : H = mc T = (100 g)(2.26 J/g°C)(25°C) = 5.650 kJ So the total H = 5.232 kJ + 33.55 kJ + 41.86 kJ + 226 kJ + 5.650 kJ = 312 kJ . 2. 1 MJ of heat is dumped into 2 kg of ice at -25°C. What is the final temperature and state (solid, liquid, or gas) of the water? Answer: First, see how much energy it takes to melt the ice: Heat ice to 0°C: H = mc T = (2000 g)(2.093 J/g°C)(25°C) = 104.65 kJ Melt ice: H = -m H fusion = -(2000 g)(-335.5 J/g) = 671.00 kJ H for this process = 104.65 + 671.00 kJ = 775.65 kJ. So we have 1000 kJ – 775.65 kJ = 224.35 kJ. So we solve for how much we can increase the temperature of the water with this amount of heat. Heat water to T f : H = mc(T f - 0°C) Tf = H/mc = 224350 kJ/(2000 g * 4.186 J/g°C) = 26.8°C 3. The phase diagram for water (shamelessly borrowed from last year’s quiz 2) is shown above. (a) What phase change(s) occur when going from 1 atm, 100 K to 1 atm, 400K? (b) From 0.1 atm, 100 K to 0.1 atm, 400K? Answer: (a) melting, then vaporization; (b) sublimation 4. For any temperature less than ~475 K, if you keep increasing the pressure on the system, what will be the eventual state of the system? Is this the same or different from most other substances?

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Worksheet 02 Key - 1 Spring 2007 CH 307 Worksheet 2 100 g...

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