1 / 54

Thermodynamics – study of energy First law of thermodynamics Energy of the universe is constant

Thermodynamics – study of energy First law of thermodynamics Energy of the universe is constant Energy can not be created or destroyed. Energy is the ability to do work or produce heat. 2 types of Energy. Kinetic energy Energy of motion E = ½ mv 2. Potential energy

suzuki
Download Presentation

Thermodynamics – study of energy First law of thermodynamics Energy of the universe is constant

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 – study of energy • First law of thermodynamics • Energy of the universe is constant • Energy can not be created or destroyed

  2. Energy is the ability to do work or produce heat. • 2 types of Energy Kinetic energy Energy of motion E = ½ mv2 Potential energy Energy of position

  3. Law of conservation of energy • Energy can not be created or destroyed but can be converted from one form to another • Ex chemical energy (gas) can be converted to mechanical and thermal energy

  4. Internal energy = E “sum” of kinetic & potential energies of all “particles” in a system Internal energy can be transferred by two types of energy flow: • Heat (q) (q = quantity of heat) • Work (w) E = q + w Change of energy in a system

  5. Measuring Energy Changes • common energy units for heat are the calorie and the joule. • Calorie – the amount of energy (heat) required to raise the temperature of one gram of water 1oC. • Joule – heat required to raise the temperature of 1 g of water by 0.24 K 1 calorie = 4.184 joules

  6. System – part of the universe on which we focus attention • Surroundings – everything else in the universe • Ex. Burning a match is the system releases heat to surroundings

  7. Endothermic – energy flows into system Exothermic – energy flows out of system

  8. C- Specific heat capacity energy required to change the mass of one gram of a substance by one degree Celsius. [joule/gram °C] • Like energy storage bank

  9. C- Specific heat capacity energy required to change the mass of one gram of a substance by one degree Celsius. [joule/gram °C] • Like energy storage bank • Water has a high specific heat capacity that means it takes a lot of heat (joules) to increase the temperature of water compared to metals (water better at storing energy than metals

  10. B. Measuring Energy Changes • To calculate the energy required for a reaction: Q = C  m t

  11. Q = C  m t How much heat is required to warm 145 g of water from 25 *C to 65 *C

  12. Q = m x C t How much heat is required to warm 145 g of water from 25 *C to 65 *C the specific heat of liquid water is 4.184 J/g *C Q = m x C t Q = 145 g x 4.184 J x (65 *C - 25 *C) g *C

  13. Q = m x C t How much heat is required to warm 145 g of water from 25 *C to 65 *C the specific heat of liquid water is 4.184 J/g *C Q = C  m t Q = 145 g x 4.184 J x (65 *C - 25 *C) g *C Q = 145 x 4.184 J x 40 Q = 24,267 joules

  14. A quantity of water is heated from 20.0 *C to 48.3 *C by absorbing 623 Joules of heat energy, what is the mass of the water? Specific heat of water is 4.184 J/g *C Q = C  m t __Q__ = m C x t

  15. A quantity of water is heated from 20.0 *C to 48.3 *C by absorbing 623 Joules of heat energy, what is the mass of the water? Specific heat of water is 4.184 J/g *C Q = C  m t __Q__ = m C x t 623 J = m 4.184 J /g *C x (48.3 *C - 20 *C)

  16. A quantity of water is heated from 20.0 *C to 48.3 *C by absorbing 623 Joules of heat energy, what is the mass of the water? Specific heat of water is 4.184 J/g *C Q = C  m t __Q__ = m C x t 623 J = m 4.184 J /g *C x (48.3 *C - 20 *C) 623 J = m 4.184 J /g *C x (28.3 *C) 5.26 g = m

  17. In a household radiator a 25.5 g of steam at 100 *C condenses (changes from gas to liquid) How much heat is released? Heat of vaporization is 2260 J/g Q = m Hvap

  18. In a household radiator a 25.5 g of steam at 100 *C condenses (changes from gas to liquid) How much heat is released? Heat of vaporization is 2260 J/g Q = m Hvap Q = 25.5 g x 2260 J/g Q = 57,630 J

  19. If an endothermic reaction absorbs 482 J how many calories are absorbed? 482 J x 1 calorie = 115 calories 4.184 J

  20. Q = m x C t How much heat is required to warm 275 g of water from 76 *C to 87 *C the specific heat of liquid water is 4.184 J/g *C Q = C  m t Q = 275 g x 4.184 J x (87 *C - 76 *C) g *C Q = 275 x 4.184 J x 11 Q = 12,657 joules

  21. In a household radiator a 1000 g of steam at 100 *C condenses (changes from gas to liquid) How much heat is released? Heat of vaporization is 2260 J/g Q = m Hvap Q = 1,000 g x 2260 J/g Q = 2,260,000 J

  22. Objectives • To consider the heat (enthalpy) of chemical reactions • To understand Hess’s Law

  23. A. Thermochemistry (Enthalpy) • Enthalpy, H– amount of heat content of a system absorbed or released at constant pressure. • H = heat = q ΔH = q  (valid for constant pressure ONLY!) That is the Enthalpy ΔH (heat content) = q (quantity of heat) The heat of formation is the enthalpy gain when a compound is CREATED from its elements at their conditions at some STP. The heat of reaction involves CHEMICAL REACTIONS OF A SUBSTANCE WITH OTHER SUBSTANCES, and is the relative enthalpy of the products compared to the reactants at some STP.

  24. Enthalpy, H– • Since H = q • And q = m x C x ΔT then… ΔH = mCΔT ΔH = mΔHvap ΔH = mΔHfus

  25. A. Thermochemistry Calorimeter • Enthalpy, H is measured using a calorimeter.

  26. Temperature- average kinetic energy Heat- flow of energy due to a temperature difference Enthalpy- heat content of a system Entropy- measure of disorder or randomness in a system

  27. Hess’s Law • the change in enthalpy is the same whether the reaction takes place in one step or a series of steps. • Example: N2(g) + 2O2(g) 2NO2(g)H1 = 68 kJ Since H= + 68 kJ then this is a endothermic reaction

  28. ΔHrxn = Heat of Reaction (represents change in enthalpy) ΔHrxn = (ΔHproducts -ΔHreactants) +ΔH = Endothermic –ΔH = Exothermic

  29. How much heat is required to raise the temperature of 445 g of zinc from 25.0*C to 74.5*C if the specific heat of zinc is 0.390 J/g*C

  30. How much heat is required to raise the temperature of 445 g of zinc from 25.0*C to 74.5*C if the specific heat of zinc is 0.390 J/g*C ΔH = mCΔT

  31. How much heat is required to raise the temperature of 445 g of zinc from 25.0*C to 74.5*C if the specific heat of zinc is 0.390 J/g*C ΔH = mCΔT ΔH = 445g x 0.390 J/g*C x 49.5*C ΔH = 8,591 Joules

  32. Determine the heat of reaction of the following equation CaCO3  CO2 + CaO Heat of formation for calcium carbonate is -1207 KJ/mol for 1 mole Heat of formation for carbon dioxide is -394 KJ/molfor 1 mole Heat of formation for calcium oxide is -635 KJ/mol for 1 mole Hreaction = SUM (Hproducts) - SUM (Hreactants)

  33. Determine the heat of reaction of the following equation CaCO3  CO2 + CaO Heat of formation for calcium carbonate is -1207 for 1 mole Heat of formation for carbon dioxide is -394 for 1 mole Heat of formation for calcium oxide is -635 for 1 mole Hreaction = SUM (Hproducts) - SUM (Hreactants) Hreaction = (-394 – 635) – (-1207)

  34. Determine the heat of reaction of the following equation CaCO3  CO2 + CaO Heat of formation for calcium carbonate is -1207 for 1 mole Heat of formation for carbon dioxide is -394 for 1 mole Heat of formation for calcium oxide is -635 for 1 mole Hreaction = SUM (Hproducts) - SUM (Hreactants) Hreaction = (-394 – 635) – (-1207) Hrxn = (-1029) – (-1207) Hrxn = +178

  35. Determine the heat of reaction of the following equation CaCO3  CO2 + CaO Heat of formation for calcium carbonate is -1207 for 1 mole Heat of formation for carbon dioxide is -394 for 1 mole Heat of formation for calcium oxide is -635 for 1 mole Hreaction = SUM (Hproducts) - SUM (Hreactants) Hreaction = (-394 – 635) – (-1207) Hrxn = (-1029) – (-1207) Hrxn = +178 ENDOTHERMIC

  36. The Heat of Formation for ALL Diatomic molecules is zero "All elements in their standard states (oxygen gas, solid carbon in the form of graphite, etc.) have a standard enthalpy of formation of zero, as there is no change involved in their formation." The problem is that we can only measure CHANGES in the Enthalpy of the system, and have no way to determine the ABSOLUTE Enthalpy. We have to define a convenient ZERO of Enthalpy for all the chemical compounds that we can make. We do this by defining the enthalpy of the Standard State of the Elements, this should cover all possible matter since all matter is made of elements we only need about 100 descriptions of Standard Elements

  37. Objectives • To understand how the quality of energy changes as it is used • To consider the energy resources of our world • To understand energy as a driving force for natural processes

  38. A. Quality Versus Quantity of Energy • When we use energy to do work we degrade its usefulness.

  39. A. Quality Versus Quantity of Energy • Petroleum as energy

  40. B. Energy and Our World • Fossil fuel – carbon based molecules from decomposing plants and animals • Energy source for United States

  41. B. Energy and Our World • Petroleum – thick liquids composed of mainly hydrocarbons • Hydrocarbon – compound composed of C and H

  42. B. Energy and Our World • Natural gas – gas composed of hydrocarbons

  43. B. Energy and Our World • Coal – formed from the remains of plants under high pressure and heat over time

  44. B. Energy and Our World • Effects of carbon dioxide on climate • Greenhouse effect

  45. B. Energy and Our World • Effects of carbon dioxide on climate • Atmospheric CO2 • Controlled by water cycle • Could increase temperature by 10oC

  46. B. Energy and Our World • New energy sources • Solar • Nuclear • Biomass • Wind • Synthetic fuels

  47. C. Energy as a Driving Force • Natural processes occur in the direction that leads to an increase in the disorder of the universe. • Example: • Consider a gas trapped as shown

  48. C. Energy as a Driving Force • What happens when the valve is opened?

  49. C. Energy as a Driving Force • Two driving forces • Energy spread • Matter spread

  50. C. Energy as a Driving Force • Energy spread • In a given process concentrated energy is dispersed widely. • This happens in every exothermic process.

More Related