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how much energy is required to change 78 grams of ice at 0 degrees celsius to steam at 165 degrees celsius

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1 Answer

1) Energy is required to break the bonds of the solid ice to change it to liquid ice. This is the Heat of fusion Hf.  For water Hf = 80 cal/g.
 
For 78g   Hf = 78g x 80 cal/g = 6240 cal
 
2) Energy to raise the temperature of the water at 0o C to 100o C
 
q = mc(Tf - Ti) where c is specific heat. For water c = 1 cal/goC
 
q = 78g x 1 cal/goC x (100oC) = 7800 cal
 
3) Energy required to change liquid water to steam at 100oC, This is the Heat of vaporization, Hv = 540 cal/g
 
Hv = 78 g x 540 cal/g = 42,120 cal
 
4) Energy required to change the temperature of steam from 100oC to 165oC
 
q = mc (Tf - Ti)   for steam c = .48 cal/goC
 
q = 78g x .48 cal/goC x 65 oC = 2434 cal
 
 
Total energy = 6240 + 7800 + 42,120 + 2434 = 58,594 cal
 
 

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