Heat, Latent Heat and Specific Latent Heat

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Presentation transcript:

Heat, Latent Heat and Specific Latent Heat Thermal Physics Heat, Latent Heat and Specific Latent Heat

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℃)

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 = 4.184 Joule (4.2 Joule) (1 Calorie = 1000 calorie)

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

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

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)

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)

An example of Solid, Liquid and Gas

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) (200-100) = 1, 507,500 J 6) QT = Q1 + Q2 + Q3 + Q4 + Q5 = 24 567 000 Joule