Calculate the amount of heat required to change 12.00g of ice at -5.00'C to water at 0.500'C?

Here's a table:


C water= 75.4 J/mol'C


C ice= 36.5 J/mol'C


C steam= 36.3 J/mol'C


^Hvap= 40.7 kJ/mol


^Hfus=6.01 KJ/mol

Calculate the amount of heat required to change 12.00g of ice at -5.00'C to water at 0.500'C?
1. Heating ice


How much heat would be required to raise 50g of ice to its melting point?


The ice temperature must be raised 10 degrees to reach 0oC.





Since the specific heat of ice is 0.50 cal/g-oC, that means that 0.50 calories is needed to raise 1g of ice 1oC. Thus, it would take 50 x 0.50 calories to raise 50g up 1oC and 10 x 50 x 0.50 = 250 cal to raise the ice to its melting point.





2. Melting ice


How much heat would be required to melt the 50g of ice?


The latent heat for melting ice is 80 cal/g. That means that 1g of ice requires 80 cal of heat to melt.





Thus, 50g requires 50 x 80 = 4000 cal to melt.





3. Heating water


How much heat is required to heat 50g of water from 0oC to its boiling point of 100oC?


Since the specific heat of water is 1.00 cal/g-oC, that means that 1.00 calorie is needed to raise 1g of water 1oC. Thus, it would take 50 x 1.00 calories to raise 50g up 1oC and 100 x 50 x 1.00 = 5000 cal to raise the water to its boiling point.





4. Boiling water


How much heat would be required to boil the 50g of water?


The latent heat for boiling water is 540 cal/g. That means that 1g of water requires 540 cal of heat to boil.





Thus, 50g requires 50 x 540 = 27000 cal to boil.





5. Heating steam


How much heat is required to heat 50g of steam from 100oC to 110oC?


Since the specific heat of steam is 0.48 cal/g-oC, that means that 0.48 calories are needed to raise 1g up 1oC. Thus, it would take 50 x 0.48 calories to raise 50g of steam 1oC and 10 x 50 x 0.48 = 240 cal to raise the temperature of the steam to 110oC.





6. Total


The total heat required to change 50g of ice at -10oC to steam at 110oC is:





250 + 4000 + 5000 + 27000 + 240 = 36490 cal.
Reply:First convert g to moles.


12.00 g x (1 mol ice/18.00 g ice) = 0.6667 mol (Note: Ice is H2O, mw 18.00)


warm the solid ice at -5.00oC to 0oC:


Heat for temp change = mol x (T2-T1) x heat capacity


0.6667 mol ice x (0o -(-5o) x 36.5 J/mol oC = 121.7 J or 0.1217 kJ


Next, melt the ice at 0oC: Heat = mol x C ice


0.6667 mol ice x 6.01 x kJ/mol = 4.007 kJ


Then warm the liquid water from 0oC to 0.500oC:


0.6667 mol liquid water x (0.500oC - 0oC) x75.4 J/mol oC =


25.13 J or 0.02513 kJ


Total heat: 0.217 kJ +4.007 kJ + 0.025 kJ = 4.249 kJ