1 / 14

Thermodynamics

Thermodynamics. A Garvey/Ziemba Production. Specific Heat Capacity. The Heat (energy) required to produce a certain temperature per gram of material Specific Heat = heat supplied (mass of object) (Temp Change)

oliver-gill
Télécharger la présentation

Thermodynamics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Thermodynamics A Garvey/Ziemba Production

  2. Specific Heat Capacity • The Heat (energy) required to produce a certain temperature per gram of material • Specific Heat = heat supplied (mass of object) (Temp Change) OR C = q/(m)(∆T) -> q = MC∆T T is in Kelvin, mass in grams, q in Joules. Specifics Heats are often given in problems

  3. Changing States Calculated by Energy = Heat of Fusion/Vaporization x mass of substance This is so easy even JC can do it!! Lets try it!

  4. Sample Problemo • How much heat is required to warm 500g of a solid teacher at -50oC to steam at 200oC? (Teachers are made only of H2O) heat of fusion = 333.5J/g, Vaporization = 2256J/g. specific heat capacities, in order of solid, liquid, gas, hinton, are 2.1, 4.2, and 2.0 J/g x K Lets go!

  5. We can do this….together! • It starts at -50oC and goes to 0o, a change of 50oK • So (500g)(2.1J/gxK)(50o) = 52,500 J • Next for Ice -> water, (500g)(333.5J/g) = 166,750J • Water at 0o to 100o -> (500g)(4.2J/gxK)(100oK) = 210,000J • Now for Water -> Steam (500g)(2256J/g) = 1,128,000J !! • Finally, Steam at 100oC to 200oC -> (500g)(2.0J/gxK)(100oK)=100,000J • Add them up -> 52,500 + 166,750 + 210,000 + 1,128,000 + 100,000 and we get…. • 1657kJ of Death!!!

  6. What we just did looks like this, with 5 steps

  7. Enthalpy change…(BORING!!!!) • The heat transferred into or out of a system (constant pressure) • Enthalpy change is Hproducts - Hreactants • Just remember products minus reactants. It works every time, no matter what unit you are studying. Don’t listen to Mr. Hinton telling me im wrong. • Enthalpies of reaction are the same!!! Sum of the products enthalpys minus sum of the reactants…every time, it works….

  8. The Hess Family Fund • Also know as Hess’s Law • Basicly, if you add reactions, the sum of the reaction’s ∆H’s is the new reaction’s ∆H • If you reverse an equation, the sign on ∆H must change • If you multiply equation, multiply ∆H by same number • Helpful Hint: for all reactions, g = 9.8m/s2

  9. Random

  10. Entropy, Free Energy • The measure of Disorder (symbol is S) • !!! ∆S is Sum of S products - Sum of S reactants • G stands for Gibbs Free Energy, another lovely Thermodynamic Function • ∆G = ∆H - T ∆S • This is Gibb’s Lovely Free Energy Equation. It is (so says the book) very important

  11. Little More Free Energy • ∆G(reaction) = Sum∆G(products) - Sum ∆G reactions • Well I know im surprised • I’d also like to take this time to remind you all the exothermic is a negative ∆H and endothermic is a positive ∆H • Better Late than never, you know

  12. Ther MO and the E constants • We all know the old saying • “∆G = -RT ln K!” • R is 8.314 • K is the THERMODYNAMIC Equilibrium CONSTANT • But hey! • If K>1 Products are favored in a reaction • If K = 1 (Rare) Its at equilibrium • If K < 1 Reactants are favored. • You, Zev and Pat, are never favored.

  13. The Last tired, slide. • Give us a good grade • Give John money • All your Base are belong to the Chemistry Department, in containers kept apart from Acids and Nick’s small thieving hands

More Related