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Units of Energy

Units of Energy. James Joule 1818-1889. 1 calorie = heat required to raise temp. of 1.00 g of H 2 O by 1.0 °C. 1000 cal = 1 kilocalorie = 1 kcal 1 kcal = 1 Calorie (a food “ calorie ” ) SI units for energy: joule (J) 1 cal = exactly 4.184 J. Energy & Chemistry. Some Basic Principles

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Units of Energy

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  1. Units of Energy James Joule 1818-1889 1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 °C. 1000 cal = 1 kilocalorie = 1 kcal 1 kcal = 1 Calorie (a food “calorie”) SI units for energy: joule (J) 1 cal = exactly 4.184 J

  2. Energy & Chemistry Some Basic Principles • Energy is defined as the ability to do work. i.e. it can be used to apply a force to move something through a distance. Heat is a form of energy.

  3. Energy has many forms • Kinetic Energy is associated with motion • Potential energy (stored energy that results from an object’s position). The law of conservation of energy: energy can neither be created nor destroyed. It can only be found in one of the forms above. Energy can be converted from one type into another using processes or machines.

  4. Energy & Chemistry Some Basic Principles • Chemical energy is stored as potential energy in bonds. • This energy can go into or out of bonds in the forms of: • Heat Electricity Light • Temperature is a measure of the average kinetic energy of molecules. Heat (q) is NOT the same as temperature.

  5. Burning peanuts supply sufficient energy to boil a cup of water. Burning sugar (sugar reacts with KClO3, a strong oxidizing agent) Energy & Chemistry

  6. Thermal Equilibrium • Energy transfer as heat will occur spontaneously from an object at a higher temperature to an object at a lower temperature. • Transfer of energy as heat continues until both objects are at the same temperature and thermal equilibrium is achieved. • At thermal equilibrium, the object with a temperature increase has gained thermal energy, the object with a temperature decrease has lost thermal energy.

  7. Heat Capacity When heat is absorbed or lost by a body, the temperature must change as long as the phase (s, g or l) remains constant. The amount of heat (q) transfer related to an object and temperature is given by: q = heat lost or gained (J) C = Heat Capacity of an object T = Tfinal  Tinitial is the temperature change (°C or K)

  8. Heat & Specific Heat Capacity  When heat is absorbed or lost by a body, the temperature must change as long as the phase (s, g or l) remains constant.  The amount of heat (q) transfer per unit mass of a substance is related to the mass and temperature by: m = mass of substance (g) q = heat lost or gained (J) C = the Specific Heat Capacity of a compound T = Tfinal  Tinitial is the temperature change (°C or K)

  9. Does it matter if we calculate a temperature change in Kelvin or degrees C? No. let Tin = 25.0 °C and Tf = 50.0 °C Recall that T = Tf –Tin 50.0 °C = 50.0 °C + 273.2 = 323.2 K Tf = –Tin =  25.0 °C =  (25.0 °C + 273.2) =  298.2 K = 25.0 °C T T = 25.0 K The are the same!

  10. Determine the final temperature of a 25.0 g block of metal that absorbs 255 cal of energy. The initial temperature of the block was 17.0 C The specific heat capacity of the metal is: 32.7 °C rearranging: +255 cal Tf = + 17.0 °C = 25.0 g Tf > Tin as expected

  11. Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at initially at 21.0 °C. • At thermal equilibrium, the water and iron are both at 23.1 °C • What is the specific heat capacity of the metal?

  12. Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at initially at 21.0 °C. • At thermal equilibrium, the water and iron are both at 23.1 °C • What is the specific heat capacity of the metal? Solution:

  13. Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at initially at 21.0 °C. • At thermal equilibrium, the water and iron are both at 23.1 °C • What is the specific heat capacity of the metal? Solution:

  14. Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at initially at 21.0 °C. • At thermal equilibrium, the water and iron are both at 23.1 °C • What is the specific heat capacity of the metal? Solution:

  15. Energy & Changes of State • When matter absorbs heat, its temperature will rise until it undergoes a Phase Change. • The matter will continue to absorb energy, however during the phase change its temperature remains constant: Phase changes are “Isothermal” processes.

  16. Energy Transfer & Changes of State • Some changes of state (phase changes) are endothermic: • When you perspire, water on your skin evaporates. • This requires energy. • Heat from your body is absorbed by the water as it goes from the liquid state to the vapor state, as a result you cool down. + energy

  17. Energy Transfer & Changes of State • Some changes of state (phase changes) are exothermic: • When it is muggy outside, water condenses on your skin. • This releases energy. • Heat from condensation is absorbed by your skin as water in the vapor state coverts to the liquid state. • As a result you feel hot. + energy

  18. Heating/Cooling Curve for Water Note the isotherms at each phase transition.

  19. Heat & Changes of State • The energy associated with a change of state is given by the Enthalpy or Heat of the phase change. • Since there is no temperature change associated with the process, the units are most often in J/g or J/mol. Sublimation: subH > 0 (endothermic) Vaporization: vapH > 0 (endothermic) Melting or Fusion: fusH > 0 (endothermic) Deposition: depH < 0 (exothermic) Condensation: conH < 0 (exothermic) Freezing: freH < 0 (exothermic) Where H refers to the “Heat” of a phase change

  20. Energy Change Calculations Heating & Cooling: q(heating or cooling)= Heat absorbed or lost in a Phase change: q(phase change) = (n = moles or grams)

  21. +333 J/g +2260 J/g Problem: What quantity of heat is required to melt 500. g of ice at 0 °C then heat the resulting water to steam at 100 °C? melt the ice  form liquid water at 0 °C  heat the water to 100 °C  boil water  fusH  Cwater  VapH

  22. Problem: What quantity of heat is required to melt 500. g of ice at 0 °C then heat the resulting water to steam at 100 °C? Equations: qphase change = mH qheat = mCT

  23. Cwater fusH vapH Problem: What quantity of heat is required to melt 500. g of ice at 0 °C then heat the resulting water to steam at 100 °C? Ice H2O(s) (0 °C) water H2O(l) (0 °C) water H2O(l) (100 °C) steam H2O(g) (100 °C)

  24. melt ice boil water heat water Cwater fusH vapH Problem: What quantity of heat is required to melt 500. g of ice at 0 °C then heat the resulting water to steam at 100 °C? Ice H2O(s) (0 °C) water H2O(l) (0 °C) water H2O(l) (100 °C) steam H2O(g) (100 °C) qtotal =

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