Question

How much energy is required to change a 44.0-g ice cube from ice at -11.0

Answer 1

1. Heating ice
How much heat would be required to raise 44g of ice to its melting point?
The ice temperature must be raised 11 degrees to reach 0^oC.

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 1^oC. Thus, it would take 44 x 0.50 calories to raise 44g up 1^oC and 11 x 44 x 0.50 = 242 cal to raise the ice to its melting point.

2. Melting ice
How much heat would be required to melt the 44g 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, 44g requires 44 x 80 = 3520 cal to melt.

3. Heating water
How much heat is required to heat 44g of water from 0^oC to its boiling point of 100^oC?
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 1^oC. Thus, it would take 44 x 1.00 calories to raise 44g up 1^oC and 100 x 44 x 1.00 = 4400 cal to raise the water to its boiling point.

4. Boiling water
How much heat would be required to boil the 44g 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,44g requires 44 x 540 = 23760 cal to boil.

5. Heating steam
How much heat is required to heat 44g of steam from 100^oC to 109^oC?
Since the specific heat of steam is 0.48 cal/g-^oC, that means that 0.48 calories are needed to raise 1g up 1^oC. Thus, it would take 44 x 0.48 calories to raise 44g of steam 1^oC and 11 x 44 x 0.48 = 232.32 cal to raise the temperature of the steam to 109^oC.

6. Total
The total heat required to change 44g of ice at -11^oC to steam at 109^oC is:

242+3520+4400+23760+232.32 = 32154.32 cal.

1 calorie = 4.18400 joules

4.184 x 32154.32 = 134,533.675 joules