How much heat in kj is evolved in converting 1.00 mol of steam at 145.0 C to ice at -55.0 C

How much heat in kj is evolved in converting 1.00 mol of steam at 145.0 C to ice at -55.0 C? The heat capacity of steam is 2.01 j.gC and of ice is 2.09 j/gc

with sig figs its comes to a total of 58.0 KJ

you will also need the heat of crystallization and condensation along with the specific heat of liquid water: -333 J/g, -2260 J/g, and 4.184 J/gC respectively

also while i was at it i checked you constants as well and they are marginally different from mine: (steam) 2.06 J/gC, and (ice) 2.11 J/gC ......... but they are close enough to not make a difference here

heres the math:

(45 + 100 + 55 = 200 which is the total change in temperature, the negative in the math work shows the direction the temperature is going)

(-45 * 2.06) + (-2260) + (-100 * 4.184) + (-333) + (-55 * 2.11) = -3220.15 J/g

H_2_O has 18 g/mol --------> 1 mol * 18 g/mol = 18 g of water

18 * -3220.15 = -57962.7 J -------> /1000 = 57.9627 KJ ---------> 58.0 KJ

I have a detailed look at a problem like yours here:

http://www.chemteam.info/Thermochem/Time…

My problem goes from -10 to 120 while yours goes from 145 down to -55. So, just reverse yours and go from -55 to 145 and follow the pattern in my example.

If you back up to the thermochemistry menu:

http://www.chemteam.info/Thermochem/Ther…

and look under problem sets, you'll see some links that take you to a variety of solved examples.