Chemistry
posted by Ryan on .
Steam at 100°C was passed into a flask containing 250 g of water at 21°C, where the steam condensed. How many grams of steam must have condensed if the temperature of the water in the flask was raised to 76°C? The heat of vaporization of water at 100°C is 40.7 kJ/mol and the specific heat is 4.18 J/(g·°C).
I use the formula
q=m*specific heat*delta T
250*4.18*(7621) = 57.475KJ
Hvap/g H20
40.7/18 = 2.26KJ/g
q=Hvap/gh20
57.475KJ/2.26KJ/g = 25.4g (answer)
It keeps telling me the answer is wrong. Can you please tell me what I may be doing wrong in this problem?

Thank you for showing your work. I think you have failed to take into account the added water from the steam AND that the final T is NOT 100 C.
So I think you have three items at play here. Steam condenses at 100 to liquid water (which adds to the mass of the water) and that heat is Hvap x mass steam. Mass steam is the unknown here.
Then you have the mass of the water from the steam x sp.h. x delta T (which is TfinalTinitial = 76100). Finally you have the original 250 g H2O being heated up from 21 to 76 (or TfTi = +55). The + signifies that heat is being added. The steam condenses to give off heat so that portion of the equation is negative and the 100 water then cools to 76 (TfTi=76100=24) so it is negative also. I hope this helps.