My name is Amanda and I am currently enrolled in CHEM II. I havenot taken Chem s

Question

My name is Amanda and I am currently enrolled in CHEM II. I havenot taken Chem since fall of 2001. I am having trouble answeringthe below question: The heat capacity of liquid water is 4.18 J/g· oC, and the heat of vaporization is 40.7kJ/mol. How many joules of heat must be provided to convert onegram of liquid water at 67 oC into one gram of steam at100 oC? The example equation given is below: In P = ( -H vap / RT ) + C So far, this is what I have, butI am not understanding where in the world C comes from. R isconstant at 8.314 J/K. C is constant too, but where do I get thenumbers to plug into C? Is this what my equation is supposed tolook like? In P = (40.7/8.314)(1/T (?)) I do not understand how Ifigure this question out. There is not an example in my book thatis similar to this problem. Please help! I am so confused! If youcould, will you please give me a step by step breakdown of theproblem so I can do additional problems like this in the futurewithout help? I would greatly appreciate it! Thank you so much inadvance!!

Explanation / Answer

Q = cp m Tliq + m Hvap

where m= mass in grams,

Tliq the temp change undergone by theliquid,

cp the specific heat capacity of the liquid,

Hvap the heat of vaporization

There would be additional terms if the vapor were heated beyondthe boiling point, or ice were present, etc, etc

In any case, you have cp = 4.18 J/g,   Hvap = 40.7 kJ/mol =40.7/18 kJ/g, and m=1g.

Q  =    cp    m     Tliq    +    m   Hvap

      = (4.18) (1) (100-67) +   (1) (22600) = 2400 J or 2.40kJ

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