Heating  Curve Answers
1.	 

qt = q1 + q2 + q3 + q4 + q5
     = mc∆T + nhfus + mc∆T + nhvap + mc∆T
     = (1.0 kg)(2.31 kJ/kgoC)(35.7oC) + (1000 g)/(44.05 g/mol)(5.02 kJ/mol) + (1.0 kg)(2.720 kJ/kgoC)(197.2oC)   
            + (1000 g)/(44.05 g/mol)(45.61 kJ/mol) + (1.0 kg)(2.310 kJ/kgoC)(71.6oC)
     = 1921.4 kJ

2.	 

qt = q1 + q2 + q3 + q4 + q5
    = mc∆T + nhvap + mc∆T + nhfus + mc∆T  
    = (75.0 g)(0.850 J/goC)(19.9oC) + (75.0 g)/(84.18 g/mol)(9.95 kJ/mol) + (75.0 g)(1.16 J/goC)(74.6oC) +
           (75.0 g)/(84.18 g/mol)(22.3 kJ/mol) + (75.0 g)(1.6 J/goC)(105.5oC)
   = 49.15 kJ
3.	 
System – CH4
   m = 
   M = 16.04 g/mol
   ∆Hcomb = -890 k/mol

   exothermic, therefore ∆H = - q
  
  ∆H = - 30.88 kJ = n∆Hcomb 
   (m)/(16.04 g/mol) = -30.88 kJ

   m = 0.557 g
	Surrounding – C25H52
    m = 150.0 g
    M = 352.77 g/mol
    ∆hfussion = 72.62 kJ/mol

q =  n∆hfussion = (150.0 g)/(352.77 g/mol)(72.62 kJ/mol)
    = 30.88 kJ

It will take 30.88 kJ to melt the wax 	

4.	   

q = q1 + q2
    = mc∆T + n∆hfussion = 501.65 kJ

∆H = - q (exothermic)
      = n∆Hrxn = - q
∆Hrxn =  - 31.3 kJ/mol
  
5.	m = 625.89 g