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Determination of Specific Heat for a Metal

Determination of Specific Heat for a Metal. 4. 3. T f. 2. 1. T i. Lines 1 and 4 are horizontal, 2 is vertical, 3 follows the experimental points until intersects 2. RVCC Fall 2009 CHEM 103-1&2 – General Chemistry I. Chapter 6: Energy and Chemical Reactions.

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Determination of Specific Heat for a Metal

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  1. Determination of Specific Heat for a Metal 4 3 Tf 2 1 Ti Lines 1 and 4 are horizontal, 2 is vertical, 3 follows the experimental points until intersects 2.

  2. RVCC Fall 2009 CHEM 103-1&2 – General Chemistry I Chapter 6:Energy and Chemical Reactions Chemistry: The Molecular Science, 3rd Ed. by Moore, Stanitski, and Jurs

  3. Thermodynamics • Thermodynamics: “heat changes”, science of heat, work and the transformations of one to the other • Thermochemistry: the study of energy changes that occur during physical processes and chemical reactions

  4. Energy Energy:the capacity to do work (w = -F×d) • Kinetic Energy is energy by virtue of motion. (mechanical, thermal, electrical…) • EK = cT = ½mv2 • Potential Energy is energy by virtue of position. (gravitational, electrostatic, chemical…) • EP = magh

  5. Energy Units 1 Joule (J) = ½ (2 kg) (1.0m/s)2 1match ~ 2000J or 2kJ 1 cal = 4.184 J 1 calorie= “the energy needed to raise the temperature of 1 gram of water from 14.5-15.5˚C” 1 Cal (dietary calorie) = 1000 cal = 1 kcal

  6. Units of Energy The label on this Australian packet of Equal indicates that it supplies 16 kJ of nutritional energy. How many “American” Calories are in this packet of Equal? 1 cal = 4.184 J 1Cal = 1000 cal

  7. Energy The Law of Conservation of Energy (First Law of Thermodynamics): Energy may be converted from one form to another, but the total quantities of energy remain constant. We will be studying the transfer of energy from one form or one place to another. How is energy transferred?

  8. Conservation of Energy – 1st Law Gravitational potential energy being converted into kinetic energy. heat work A single Fritos chip burns in O2 The chemical potential energy stored in food (chemical bonds) being converted into thermal (kinetic) energy.

  9. Work • Workis one process that transfers energy to an object. • For example: • separate a Na+ from a Cl- – increases the potential energy of the object • throwing a baseball – increases the kinetic energy of an object

  10. Heat Heat is another process that transfers energy to an object. Heat refers to the energy transfer that occurs whenever two samples of matter at different temperatures (thermal or kinetic energy) are brought into contact. Energy always transfers from the hotter sample to the cooler sample until both are at the same temperature (in thermal equilibrium).

  11. Heat All matter consists of nanoscale particles that are constantly in motion (KMT – Ch.1). Higher temperature  faster motion of particles  more thermal energy Raising the temperature of a sample increases the thermal (kinetic) energy of that sample. So… does CO2(g) and C12H22O11 (s) at room temperature have the same kinetic energy?

  12. Specific Heat Capacity The specific heat capacity (c) is the heat required to raise the temperature of one gram of a substance by one degree Celsius. q - heat absorbed or released c – specific heat for a substance DT – change in temperature

  13. Specific Heat Metals have low specific heats. Water has high specific heat. (4.184 J/gC or 1.000 cal/gC)

  14. Example 1 If a 10.0 g sample of each substance below has 250 J applied to it. Which substance will have the greatest increase in temperature? iron (c = 0.46 J/gC) water (c = 4.184 J/gC) copper (c = 0.39 J/gC) aluminum (c = 0.92 J/gC)    - constant

  15. Example 2 – solve for c Calculate the specific heat of a substance if 2.35 kJ of heat is needed to raise the temperature of 200. g of this substance from 19.0ºC to 98.0ºC  ?     

  16. Example 3 – solve for qloss How much energy will be loss when 500.0 g of iron cools from 55°C to 22°C? cFe = 0.451 J g-1 °C-1. How much energy will be loss when 500.0 g of gold cools from 55°C to 22°C? cAu = 0.128 J g-1 °C-1. -7400J or -7.4kJ -2100J or 2.1kJ

  17. Example 4 – solve for T What will be the final temperature of a 50.00 g silver ring at 37.0 C that gives off 25.0 J of heat to its surroundings (c = 0.235 J/g C)?    ? ? 

  18. Heat Transfer (gain = loss) The hot steel bar transfers energy to the cool water until the two are at the same temperature (in thermal equilibrium).

  19. LAB! - Transfer of Kinetic Energy • qloss,Fe = qgain,H2O • mFe cFeTFe = mH2O cH2O TH2O

  20. Example 5 : -qloss = qgain , solve c A 19.6 g sample of metal was heated to 61.67 ºC and placed in 26.7 g of water in a coffee cup calorimeter. The temperature of water increased from 25.00 ºC to 30.00 ºC. What is the specific heat of the metal?

  21. Example 6 -qloss = qgain , solve ∆T A 200. g block of Fe heated in a bunsen burner flame is plunged into 1.00 kg of water (T = 23.4 °C) in an insulated container. The final equilibrium T of water and Fe is 33.4˚C. The specific heat of Fe is 0.448J/g°C. What was the temperature of the flame? heat lost by Fe = heat gained by H2O −(200.g)(0.448J/g°C)(∆Tm)= +(1000.g)(4.184 Jg-1°C-1)(33.4− 23.4°C) ∆Tm =+(1000.g)(4.184 Jg-1°C-1)(10.0°C) = -467°C −(200.g)(0.448J/g°C) ∆Tm = -467°C=33.4 °C -Ti Ti = 500.°C

  22. [From Applied Optics vol. 11, (14) 1972] The temperature of Heaven can be rather accurately computed. Our authority is Isaiah 30:26, "Moreover, the light of the Moon shall                     be as the light of the Sun and the light                     of the Sun shall be sevenfold, as the                     light of seven days."          Thus Heaven receives from the Moon as much radiation as we do from the Sun, and in addition 7*7 (49) times as much as the Earth does from the Sun, or 50 times in all. The light we receive from the Moon is one 1/10,000 of the light we receive from the Sun, so we can ignore that ... The radiation falling on Heaven will heat it to the point where i.e., Heaven loses 50 times as much heat as the Earth by radiation. Using the Stefan-Boltzmann law for radiation, (H/E)4 =50 and the temperature of the earth (300K), gives H (heaven) as 798K (525°C). The exact temperature of Hell cannot be computed ... [However] Revelations 21:8 says "But the fearful, and unbelieving ... shall                     have their part in the lake which burneth                     with fire and brimstone."         A lake of molten brimstone (sulfur) means that its temperature must be at or below the boiling point, 444.6°C. So… Heaven (525°C) is hotter than Hell (445°C)!

  23. Energy Transfer, H Enthalpy • q - heat, energy transfer (kinetic) that occurs when two samples at different temperature are brought into contact. • H = qp - enthalpy, energy transfer (potential) that occurs when a system changes physical state or undergoes a chemical reaction. stored in bonds intermolecular bonds intramolecular bonds

  24. Enthalpy, Heats of Reaction As already stated: • Anendothermic processis a chemical reaction or physical change in which heat is absorbed (qp is positive). • Anexothermic processis a chemical reaction or physical change in which heat is released (qp is negative).

  25. Phase Changes –Endothermic and Exothermic • Boiling Endothermic qp < 0 (-) net bond breaking • Condensation Exothermic qp > 0 (+) net bond breaking Which is worse? A burn from 100˚C water or 100˚C steam?

  26. Changes of State, Enthalpy Change Name value for H2O (J/g) solid → liq enthalpy of fusion (melting) 333 liq → gas enthalpy of vaporization 2260 liq → solid enthalpy of freezing −333 gas → liq enthalpy of condensation −2260 Note: ΔHfusion = − ΔHfreezing

  27. Energy Transfer, q and H no change in temperature! q H(qp) q and H

  28. Phase Changes:Freezing and Melting of Water KE q=mcT PE H KE q=mcT

  29. Temperature (°C) -50 -25 0 25 50 Water warms from 0 to 50°C Ice warms from -50 to 0°C 0 100 200 300 400 500 600 Quantity of energy transferred (J) Problem • How much energy is used in converting 5.0g of ice at -50.˚C to water at 50.˚C? ΔHfusion = 333 J/g cwater = 4.184 J/g˚C cice = 2.06 J/g˚C Ice is melting. T remains at 0°C 3200 J

  30. Enthalpy Change- Chemical Reactions Endothermic DH > 0; ( + ) net bond breaking Exothermic DH < 0; ( - ) net bond making

  31. H2(g) + Cl2(g) 2 H(g) + 2 Cl(g) 2 HCl(g) endothermic ΔH= +678 kJ/mol exothermic ΔH= -862 kJ/mol Bond Enthalpies During a chemical reaction: Old bonds break: requires E (endothermic) New bonds form: releases E (exothermic) Both typically occur: net bond breaking, -184 kJ/mol

  32. Thermochemical Equations Athermochemical equationis the chemical equation for a reaction in which the enthalpy of reaction for these molar amounts is written directly after the equation. When 2 moles of ammonia are produced, the reaction gives out (produces) 92.22 kJ of heat. ΔHo is the enthalpy change at the standard pressure of 1 bar and a specified temperature. When temperature is not specified, it is intended to be 25oC

  33. WritingThermochemical Equations In athermochemical equationit is important to note phase labels because the enthalpy change, DH, depends on the phase of the substances.

  34. WritingThermochemical Equations Exothermic reaction: Can be written as: Heat is a product. Endothermic reaction: Can be written as: Heat is a reactant.

  35. ManipulatingThermochemical Equations Rule 1: When a thermochemical equation is multiplied by any factor, the value of DH for the new equation is obtained by multiplying the DH in the original equation by that same factor. Multiply by 1/2

  36. ManipulatingThermochemical Equations Rule 2: When a chemical equation is reversed, the value of DH is reversed in sign.

  37. Example 1 What is the enthalpy change for the following reactions?

  38. Example 2 Given the following thermochemical reaction: N2(g) + 3H2(g)  2NH3(g) DH˚ = -92.22kJ Write the thermochemical reaction for: a.) Formation of 1 mol of ammonia gas b.) Decomposition of 6 moles of ammonia gas c.) Combination of 1 mole of hydrogen with a stoichiometric amount of nitrogen to produce ammonia gas

  39. a.) Formation of 1 mol of ammonia gas N2(g) + 3H2(g)  2NH3(g) DH˚ = -92.22kJ a.) 1/2 N2(g) + 3/2 H2(g)  NH3(g) DH˚ = -46.11kJ b.) Decomposition of 6 moles of ammonia gas b.) 6NH3(g)  3N2(g) + 9H2(g) DH˚ = 276.66kJ c.) Combination of 1 mole of hydrogen with a stoichiometric amount of nitrogen to produce ammonia gas c.) 1/3 N2(g) + H2(g)  2/3 NH3(g) DH˚ = -30.74kJ

  40. Heat StoichiometryInstant Cold Pack H˚ = +28.1kJ ICYHOT Instant Cold Pack contains 60.g of NH4NO3. How many kJ of heat are absorbed by the system (loss by the surroundings)? +2.1 kJ

  41. Heat StoichiometryInstant Heat Pack H˚ = -824 kJ HEAT TREAT Hand Warmer contains 2.5 g of Fe. How many kJ of heat are released by the system (gained by the surroundings)? -18 kJ

  42. Enthalpy Enthalpy is a state function – the enthalpy of a system depends only on its present state and not on the route by which it got there. This allows us to apply laboratory measurements to real-life situations. eg. The enthalpy change of a chemical reaction is the same – whether it occurs in a lab or in your body. Altitude is a state function.

  43. Hess’s Law “If the equation for a reaction is the sum of the equations for two or more other reactions, then ΔH° for the 1st reaction must be the sum of the ΔH° values of the other reactions.” “ΔH° for a reaction is the same whether it takes place in a single step or several steps.”

  44. Example It is difficult to measure ΔH for: 2 C(graphite) + O2(g) 2 CO(g) Some CO2 always forms. Calculate ΔH given: C(graphite) + O2(g) CO2(g) ΔH = −393.5 kJ 2 CO(g) + O2(g) 2 CO2(g) ΔH = −566.0 kJ Hess’s Law Use Hess’s Law to find ΔH for unmeasured reactions.

  45. 2 C + 2 O2 2 CO2ΔH° = −787.0 kJ 2 CO2 2 CO + O2 ΔH° = +566.0 kJ 2 C + O2 2 CO ΔH° = −221.0 kJ Hess’s Law Calculate ΔH for the reaction: 2C(graphite) + O2(g) → 2CO(g) Rearrange: +2[C + O2 CO2] +2(−393.5) = −787.0 −1[2 CO + O2 2 CO2] −1(−566.0) = +566.0 or: Add, then cancel: 2 C + 2 O2 + 2 CO2 2 CO2 + 2 CO + O2

  46. 2 C(s) + 2 H2O(g) CH4(g) +CO2(g) C(s) + H2O(g) CO(g) + H2(g)ΔH° = 131.3 kJ CO(g) + H2O(g) CO2(g) + H2(g)ΔH° = −41.2 kJ CH4(g) + H2O(g) CO(g) + 3 H2(g)ΔH° = 206.1 kJ Hess’s Law Determine ΔH° for the production of coal gas: Using: A B C

  47. 2 C on left, use 2 x A 2 C(s) + 2 H2O(g) 2 CO(g) + 2 H2(g)+262.6 1 CH4 on right, use −1 x C CO(g) + 3 H2(g) CH4(g) + H2O(g) −206.1 1 CO2 on right, so use 1 x B CO(g) + H2O(g) CO2(g) + H2(g) −41.2 Add and cancel: 2C + 3H2O + 2CO + 3H2 2CO + 3H2 + CH4 + H2O + CO2 Hess’s Law Want:2 C(s) + 2 H2O(g) CH4(g) + CO2(g) 2 C + 2 H2O → CH4 + CO2 ΔH = 15.3 kJ

  48. Heats of Reaction: Calorimetry Acalorimeteris a device used to measure the heat absorbed or evolved during a physical or chemical change. Whatever heat is produced during a chemical reaction, it is absorbed by the calorimeter. Whatever heat is absorbed by the chemical reaction, it will be supplied by the calorimeter Total heat, produced or absorbed remains inside the system

  49. Calorimetry When reactions take place in solution, it is easier to use a calorimeter that is open to the atmosphere. This allows for the direct measurement of H (q at constant pressure). An easy way to do this is with a coffee-cup calorimeter.

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