This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 3. Warm the water to final temp: q = 9.17x10 6 g *(4.184J/g C)*(200 C) = 767345.6 kJ required. Grand total = 3922182kJ 4. Time to run heater: 3922182kJ *(1s/40kJ)*(1hr/3600s) = 27.2hrs!!! ' H fus of benzene is 10.59 kJ/mol. I have 47g of liquid benzene at its melting point, 278.6K. If I remove 2kJ of heat energy from this sample, what is the final temperature? 47g X 1mol/78g = 0.60mol. 1. Try removing all the heat. See how many moles of benzene freeze. q = 2kJ = 10.59kJ/mol * ?mol = 0.188 moles of benzene freeze. This is LESS than the total amount we have, so, 0.6  0.188 moles = 0.411 = 32 g left as liquid. If we had arrived at a number above 0.60 mole we would have to find out how much heat is removed to freeze all the benzene, and wed need the specific heat of solid benzene to find the final temperature...
View
Full
Document
This note was uploaded on 02/13/2012 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Holcombe
 Chemistry

Click to edit the document details