How much heat (in KJ) is evolved in converting 1.00 mol of steam.

+3 votes

How much heat (in {\rm kJ}) is evolved in converting 1.00 {\rm mol} of steam at 135.0^\circ C to ice at -50.0^\circ C? The heat capacity of steam is 2.01 {\rm{J}}/{\rm{g}} \cdot ^\circ {\rm{C}} and of ice is 2.09 {\rm{J}}/{\rm{g}} \cdot ^\circ {\rm{C}}.

asked Jun 18, 2012 in General Chemistry by kistnus ~Expert~ (614 points)
    

1 Answer

+2 votes

Steam, ice, and liquid water are all just H2O.

1.00 mole steam = 18g H2O


First you cool the steam from 135 C to 100 C, a difference of 35 C:

18 g H2O x 35 C x 2.01 J/g C =  ?

Next, the steam condenses to water:

18 g H2O x 2258 J/g H2O = 40,644 J

Next, cool the water to 0C:

18 g H2O x 100 deg C x 4.184J /g-C = ? J

Next freeze the water:

18 g H2O x 333 J/g H2O = 5994 J

Next cool the ice to -45.0C:

18 g H2O x 50 C x 2.09 J/g C = ? J

Now add up all the energies to obtain final answer :

 

answered Jun 18, 2012 by Gummmys ~Expert~ (1,105 points)
edited Jun 18, 2012 by Gummmys



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