ch302 notes ch16.1 - Example H fus of benzene is 10.59...

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Example Δ H fus of benzene is 10.59 kJ/mol. I have 47g of liquid benzene at its melting point, 278.6K. If I remove 2000J of heat energy from this sample, how much liquid benzene is left? 47g X 1mol/78g = 0.60mol q = 2kJ = 10.59kJ/mol * ?mol mol = 0.188 moles of benzene freeze so, 0.6 - 0.188 moles = 0.411 = 32 g left as liquid.
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Example What will be the final temperature of 20L of water at room temperature (25.000°C) if 200g of ice at –1 ° C is dropped into it? What happens? BREAK IT INTO STEPS! 1. 200g ice at –1 ° C warms to 0C: Heat energy flows from water to ice: q (into ice) = 200g * (2.03J/g °C) * (1 °C) = 406J absorbed from warm water. 2. Warm water cools slightly as result: q (out of warm water) = -q (into ice) = - 406J If - 406J = 20L * 1000g/1L * 4.184J/g °C * Δ T this gives Δ T = 0.00485 ° C Water is now at (25 - 0.00485)° C = 24.995° C 3. 200g ice at 0 ° C melts: Heat energy flows from water to ice: q (into ice) = 200g/18(g/mol) * 6.01kJ/mol= 66777.8J absorbed from warm
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This note was uploaded on 05/01/2009 for the course CH 52410 taught by Professor Sutcliffe during the Spring '09 term at University of Texas at Austin.

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ch302 notes ch16.1 - Example H fus of benzene is 10.59...

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