​During a phase change, heat is either absorbed (endothermic) or released (exothermic)

​The heat associated with a phase change is directly related to...

• The mass , m, of the sample (units of grams)

• A constant based on the phases the change involves

  • if transition between solid and liquid phases... (a.k.a. freezing or melting)

    • Use H_f , heat of fusion

  • if transition between liquid and gas phases... (a.k.a. vaporizing or condensing)

    • Use H_v, heat of vaporization

  ​ex​ample calculations next... (with specific values for H_f and H_v included)

​Example One:

Calculating the heat absorbed when 100 g of water melts using formula q = mH_f ​

​

• m = 100 g

• heat of fusion, H_f = 334 J/g

• q = 100 g x 334 J/g = 33400 J

​Example Two:

Calculating the heat released when 100 g of water freezes using formula q = mH_f ​

​

• m = 100 g

• heat of fusion, H_f = 334 J/g

• q = 100 g x 334 J/g = 33400 J

​

Yeah... the same amount of heat absorbed when 100 g of water melts.

​

Energy is conserved... so opposite processes have equal but opposite energy change.​

​Example Three:

Calculating the heat absorbed when 100 g of water vaporizes (boils) using formula q = mH_v ​

​

• m = 100 g

• heat of vaporization, H_v = 2260 J/g

• q = 100 g x 2260 J/g = 226000 J

​

It is a very similar equation. Only the value of the constant is different. Notice that it takes more energy per gram of water to boil it than to melt it.​

​Next up... 2 problems as an exit ticket to see how you are doing with these equations...

Energy values can get quite large... so kilojoules (kJ) or even megajoules (MJ) are often used instead of the base unit Joule, J.

For example... 4500 J could be reported as 4.500 kJ