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# Chapter 16 Notes-Answers - 3 Warm the water to final temp q...

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If water’s normal BP is 100 q C, what P do you have to bet at for water to boil at say 20 q C? ' H vap = 40.65 kJ/mol ln P T 1vap = ' H vap {(1/T 2 )-(1/T 1 )} P T 2vap R Use P T 1vap = 760 torr and P T 2vap = x torr T 1 = 100 q C = 373K T 2 = 20 q C = 293K {(1/T 2 )-(1/T 1 )} = 7.320 x 10 -4 ln {760 torr / x torr } = 3.579 so x = 21.2 torr A swimming pool containing 10,000L of water is frozen solid, at -5 ° C. (Assume its one of those above ground ones). A 40kW heater (40kJ/s) is to be used to melt the ice and heat the pool to 20 ° C. How long does the heater have to run? (The density of ice is 0.917 g/mL). What happens? BREAK IT INTO STEPS! Mass of ice: m = 10,000 L*(1000mL/1L)*(0.917g/mL)= 9.17x10 6 g 1. The ice at –5 ° C warms to 0C: Heat energy flows to ice: q = 9.17x10 6 g*(2.03J/g ° C)*(0-(-5) ° C) = 93075.5 kJ required. 2. The ice at 0 ° C melts: Heat energy flows to ice: q = 9.17x10 6 g *(1mol/18g)*6.01kJ/mol = 3061761kJ required.
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Unformatted text preview: 3. Warm the water to final temp: q = 9.17x10 6 g *(4.184J/g ° C)*(20-0 ° C) = 767345.6 kJ required. Grand total = 3922182kJ 4. Time to run heater: 3922182kJ *(1s/40kJ)*(1hr/3600s) = 27.2hrs!!! ' H fus of benzene is 10.59 kJ/mol. I have 47g of liquid benzene at its melting point, 278.6K. If I remove 2kJ of heat energy from this sample, what is the final temperature? 47g X 1mol/78g = 0.60mol. 1. Try removing all the heat. See how many moles of benzene freeze. q = 2kJ = 10.59kJ/mol * ?mol = 0.188 moles of benzene freeze. This is LESS than the total amount we have, so, 0.6 - 0.188 moles = 0.411 = 32 g left as liquid. If we had arrived at a number above 0.60 mole we would have to find out how much heat is removed to freeze all the benzene, and we’d need the specific heat of solid benzene to find the final temperature...
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