How much energy is required to change a 25-g ice cube from ice at -25 degrees Celsius to steam at 120 celsius
1. Heating ice
How much heat would be required to raise 25g of ice to its melting
point?
The ice temperature must be raised 25 degrees to reach 0oC.
Since the specific heat of ice is 0.250 cal/g-oC, that means that
0.250 calories is needed to raise 1g of ice 1oC. Thus, it would
take 25 x 0.250 calories to raise 25g up 1oC and 25 x 25x 0.250 =
2500 cal to raise the ice to its melting point.
2. Melting ice
How much heat would be required to melt the 25g 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, 60g requires 25 x 80 = 2000 cal to melt.
3. Heating water
How much heat is required to heat 25g 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 25 x 1.00 calories to raise 25g up 1oC and 100 x 25 x 1.00 =
2500 cal to raise the water to its boiling point.
4. Boiling water
How much heat would be required to boil the25g 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, 60g requires 25 x 540 = 13500 cal to boil.
5. Heating steam
How much heat is required to heat 25g of steam from 100oC to
125oC?
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 take25
x 0.48 calories to raise 25g of steam 1oC and 25 x 25 x 0.48 = 300
cal to raise the temperature of the steam to 125oC.
6. Total
The total heat required to change 25g of ice at -25oC to steam at
125oC is:
2500 +2000 + 2500 + 13500 + 300 = 18300 cal.
1 calorie = 4.18400 joules
4.184 x 18300 = 76567.2 joules
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