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72. How much heat is necessary to convert 25.0 g of ice at -15.0oC into steam at 145.0oC?
cice = 2.06 J/goC; cwater = 4.18  J/goC; csteam = 2.02  J/goC; Lfusiton = 334 J/g; and
Lvaporization = 2260 J/g
       
A. 6690 J C. 78,400 J
B. 56,500 J D. 81,600 J
     
ice at - 15.0oC to steam at 145.0oC  

 (Break into steps in which either phase or temperature changes but not both)

 

ice at -15.0oC to ice 0.0oC
ice 0.0oC to water 0.0oC
water 0.0oC to water 100.0oC
water 100.0oC to steam 100.0oC
steam 100.0oC to steam 145.0oC

q = mcice15.0oC
q = mLfusion
q = mcwater 100.0oC
q = mLvaporization
q = mcsteam45.0oC

= 25.0 g x 2.06 J/goC x 15.0oC
= 25.0 g x 334 J/g
= 25.0 g x 4.18  J/goC x 100.0oC
= 25.0 g x 2260 J/g
= 25.0 g x 2.02  J/goC x 45.0oC

  
q = 773 J  +  8350 J  +  10500 J  +  56500 J  +  2270 J =  78,400 J

The Correct Answer is C.