Chapter 14: Liquids and Solids Energy requirements for phase changes

Learning Target You will learn to calculate energy requirements for phase changes.

Heating/Cooling curve for water

Energy Requirements It takes energy to melt and vaporize water. Remember chemical bonds are not broken in these processes. The energy required to melt 1 mol of a substance is called the molar heat of fusion. The energy required to change 1 mol of liquid to vapor is called the molar heat of vaporization.

Constants and How to Solve The molar heat of fusion for ice is 6.02 kJ/mol. The molar heat of vaporization for water is 40.6 kJ/mol. To solve for the energy to melt or vaporize a substance mass  moles  kJ (using the molar heat of fusion or vaporization value)

How to solve To solve for the energy to change the state of a substance you break the process into 3 steps: The change in physical state q = s x m x ΔT (this is from Ch. 10!!!) The vaporization and/or fusion process mass  moles  kJ (using the molar heat of vaporization value) The total energy sum up your steps

Examples How much heat is required to melt 50.0 g of ice at 0°C? (the molar heat of fusion for water is 6.02 kJ/mol) How much heat is necessary to completely vaporize 75.0 g of ice at 0°C? The molar heat of fusion for water is 6.02 kJ/mol. The specific heat capacity of liquid water is 4.184 J/g°C. The molar heat of vaporization for water is 40.6 kJ/mol.

Example Calculate the energy released when 15.5 g of ice freezes at 0°C. The molar heat of fusion of ice is 6.02 kJ/mol Answer: 5.18 kJ Mass to moles to kJ

Example Calculate the energy required to vaporize 35.0 g of water at 100°C. The molar heat of vaporization of water is 40.6 kJ/mol. Answer: 78.9 kJ Mass to mol to kJ

Example Calculate the energy required to melt 12.5 g of ice at 0°C and change it to water at 25°C. The specific heat capacity of liquid water is 4.18 J/g°C and the molar heat of fusion of ice is 6.02 kJ/mol. Answer: 5.48 kJ Mass to mol to kJ Q = m * s * change in temperature

Example Calculate the energy required to heat 22.5 g of liquid water at 0°C and change it to steam at 100°C. The specific heat capacity of liquid water is 4.18 J/g°C and the molar heat of vaporization of water is 40.6 kJ/mol. Answer: 60.1 kJ Q = m * s * change in temperature Mass to mol to kJ

Example Calculate the total energy required to melt 15 g of ice at 0°C, heat the water to 100. °C, and vaporize it to steam at 100. °C. Hint: Break the process into three steps and then take the sum. 45 kJ Q = m * s * change in temperature Mass to mol to kJ (twice)

Recap/Quiz-ANSWER ON OWN! Create the heating/cooling curve for water How much heat is required to melt 50.0 g of ice at 0°C? (the molar heat of fusion for water is 6.02 kJ/mol) How much heat is necessary to completely vaporize 75.0 g of ice at 0°C? The molar heat of fusion for water is 6.02 kJ/mol. The specific heat capacity of liquid water is 4.184 J/g°C. The molar heat of vaporization for water is 40.6 kJ/mol. (HINT: 3 #’s to add)

Answers On board 16.7 kJ 225 kJ