1. Lecture 23: Heat l Internal Energy l Heat l Specific Heat l Latent Heat l Phase Diagrams

2. Internal Energy l Energy of all molecules including: è Random motion of individual molecules » = 3/2 k T for ideal gas »Vibrational energy of molecules and atoms è Chemical energy in bonds and interactions l DOES NOT INCLUDE è Macroscopic motion of object è Potential energy due to interactions w/ other objects

3. Heat l Definition: Flow of energy between two objects due to difference in temperature è Note: similar to WORK è Object does not “have” heat (it has energy) l Units: calorie è Amount of heat needed to raise 1g of water 1ºC è 1 Calorie = 1000 calories = 4186 Joules

4. Specific Heat l Heat adds energy to object/system. l If system does NO work then:  Heat increases internal energy. Q =  U è Heat increases temperature! è Heat required to increase temperature depends on amount of material (m) and type of material (c). Q = c m  T è Q = heat è c = specific heat è m = mass   T = change in temperature

5. Specific Heat: Ideal Gases l Monatomic Gas (single atom): è All energy is translational kinetic energy. è At constant volume, work = 0.  Q =  K tr = 3/2 nR  T è C V = 3/2 R = 12.5 J/(K mole) l Diatomic Gas (two atoms): è Also have rotational energy. è C V = 5/2 R = 20.8 J/(K mole)

6. Latent Heat l As you add heat to water, the temperature increases for a while, then it remains constant, despite the additional heat! l Q = m L l Latent Heat: L [J/kg] is heat which must be added (or removed) for material to change phase (solid- liquid or liquid-gas). T Q added to water water: temp rises water changes to steam (boils) steam: temp rises 100 o C Latent Heat

7. Summary l Heat is FLOW of energy è Heat will increase internal energy. l Specific Heat  Q = c m  t è Monatomic IDEAL Gas: C V = 3/2 R è Diatomic IDEAL Gas: C V = 5/2 R l Latent Heat è Heat associated with change in phase è Q = m L

8. Example l How much heat is needed to change 1.8 grams of ice at -10ºC to steam at 100ºC? »c ice = 2.100 J/g/ºC »c water = 4.186 J/g/ºC »L fusion = 333.7 J/g »L vaporization = 2256 J/g We will need to use both equations for this problem: Q = c m  T to raise the temperature of the ice from -10ºC to 0ºC Q = m L to melt the ice to water Q = c m  T to raise the temperature of the water from 0ºC to 100ºC Q = m L to boil the water to steam

9. Example l How much heat is needed to change 1.8 grams of ice at -10ºC to steam at 100ºC? »c ice = 2.100 J/g/ºC »c water = 4.186 J/g/ºC »L fusion = 333.7 J/g »L vaporization = 2256 J/g Set up an expression for the total heat needed: Q = c ice m  T ice + m L fusion + c water m  T water + m L vaporization Solve for the total heat needed: Q = 37.8 J + 600.66 J + 753.48 J + 4060.8 J = 5452.74 J

10. Phase Diagrams