1. Heat, Latent Heat and Specific Latent Heat
Thermal Physics
Heat, Latent Heat and Specific Latent Heat

2. Heat Heat is basically a state of temperature
It can be used for both, the cold and hot
It can be applied to change the phases of matter (S, L, G)
Boiling temperature of water is 100 degree celsius (100℃)
Melting temperate of ice is 0 degree celsius
(0℃)

3. Heat Formula and Unit
Heat, Q = m L (Joule) (Phase Change) (Latent State)
Heat, Q = mcΔT (Joule) (Temp Change)
Latent Heat of Fusion, Lfus = 3.33 x 105 J/kg (S→L)
Latent Heat of Vaporization, Lvap = 22.6 x 105 J/kg (L→G)
mass, m in kg
specific heat capacity, c (water = 4186 J/kg℃)
specific heat capacity, c (ice = 2100 J/kg℃)
specific heat capacity, c (steam = 2010 J/kg℃)
Unit SI , Heat (Joule) [Energy or Work or Workdone (Joule)]
1 calorie = Joule (4.2 Joule) (1 Calorie = 1000 calorie)

4. intermolecular forces, rapid random motion high temp low pressure
Phases of matter
Gas - very weak
intermolecular forces,
rapid random motion
high temp
low pressure
Liquid - intermolecular
forces bind closest neighbours
Solid - strong
intermolecular forces
low temp
high pressure

5. Latent Heat Some other technical terms are either hidden heat or
transformational heat
Latent heat remains same throughout a process but
paves way for complete
transformational phases of
S, L and G

6. Latent Heat of Melting (or Fusion)
The heat energy that supplied to a solid, as to change it into
liquid state, where the
temperature remains
unchanged
(Ice begin to melt at zero degree
celsius while water begin to
freeze at the same temp)

7. Latent Heat of Vaporization
The heat energy that supplied to a liquid, as to change it into
gas state, where the
temperature remains
unchanged
(Water begin to vaporize at 100℃
celsius while gas begin to
condense at the same temp)

8. An example of Solid, Liquid and Gas

9. How much energy is required to heat 7
How much energy is required to heat 7.5 kg of ice from - 30℃ to steam at 200℃
1) Q1 = mcΔT = 7.5 (2100) (0-(-30)) = 472,500 J 2) Q2 = mL = 7.5 (3.33 x 105) = 2, 497,500 J 3) Q3 = mcΔT = 7.5 (4186) (100-(0)) = 3, 139,500 J 4) Q4 = mL = 7.5 (22.6 x 105) = 16, 950,000 J 5) Q5 = mcΔT = 7.5 (2010) ( ) = 1, 507,500 J 6) QT = Q1 + Q2 + Q3 + Q4 + Q5 = Joule