Asked Nov 17, 2019

Calculate the change in heat when 22.50  g of water vapor (steam) at 100.0°C condenses to liquid water and then cools to 13.50 °C.


Expert Answer

Step 1

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22.50 g Mass of water vapor Heat of vaporization of water =40.79 kJ/mol 1 mol -40.79 kJ Amount of heat released, q 22.50 gx X 18.02 g 1 mol 50.93KJ = -50930J

Step 2

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Mass of water= 22.50g J Heat capcity of liquid water, s = 4.184 g.C Initial temperature, t, -100 °C Final temperature, t, = 13.50 °C Amount heat lost in the process 92 msAT J = 22.50g x 4.184 g.C -x(13.50 C-100'c) =-8143.11J


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