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Thermodynamics

Thermodynamics. Ch 10 Energy Sections 10.4-10.6. Thermodynamics. The Law of Conservation of Energy is also known as The 1st Law of Thermodynamics . . It can be stated as “the energy of the universe is constant.”. Internal Energy (E) = kinetic energy + potential energy. Δ E = q + w

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Thermodynamics

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  1. Thermodynamics Ch 10 Energy Sections 10.4-10.6

  2. Thermodynamics The Law of Conservation of Energy is also known as The 1st Law of Thermodynamics. It can be stated as “the energy of the universe is constant.” Internal Energy (E) = kinetic energy + potential energy ΔE = q + w Δ (delta) means a change q = heat absorbed by the system w = work done on the system

  3. Units for Measuring Heat The Joule is the SI system unit for measuring heat: The calorie is the heat required to raise the temperature of 1 gram of water by 1°Celsius 1 cal = 4.184 J

  4. Energy & Temperature of Matter The amount the temperature of an object increases depends on the amount of heat added (q). • If you double the added heat energy the temperature will increase twice as much. The amount the temperature of an object increases when heat is added depends on its mass • If you double the mass it will take twice as much heat energy to raise the temperature the same amount.

  5. Converting Calories to Joules How many calories of energy correspond to 28.4 J? • 28.4 J x 1 cal = 6.79 cal • 4.184 J • 34.8 cal x 4.184 J= 146 J • 1 cal • 47.3 J x 1 cal = 11.3 cal • 4.184 J Express 34.8 cal of energy in units of joules. Express 47.3 J of energy in units of calories.

  6. Specific Heat The amount of heat required to raise the temperature of one gram of substance by one degree Celsius.

  7. Calculations Involving Specific Heat OR c = Specific Heat Q = Heat lost or gained T = Temperature change m = Mass

  8. The amount of heat required to raise the temperature of one gram of substance by one degree Celsius. Specific Heat

  9. J Specific heat of water = 4.184 ° g C Calculate the amount of heat energy (in joules) needed to raise the temperature of 7.40 g of water from 29.0°C to 46.0°C. c = Specific Heat Q = Heat lost or gained T = Temperature change m = Mass Mass = 7.40 g Temperature change = 46.0°C – 29.0°C = 17.0°C

  10. J Specific heat of gold = 0.13 ° g C A 1.6 g sample of metal that appears to be gold requires 5.8 J to raise the temperature from 23°C to 41°C. Is the metal pure gold?. c = Specific Heat Q = Heat lost or gained T = Temperature change Mass = 1.6 g Temperature change = 41.0°C – 23.0°C = 18.0°C m = Mass The metal cannot be pure gold!

  11. Calculate the joules of energy required to heat 454g of water from 5.4°C to 98.6°C ? Remember, 4.184 J of energy is required to change the temperature of each gram of water by 1°C, so……… • 4.184 J x 454 g x 93.2°C = 1.77 X 105 J g °C

  12. Determine the amount of energy as heat that is required to raise the temperature of 1.0 L of water from 25.0 C to boiling. (in Joules? in Calories?) if 1 mL of H2O = 1 g of H2O then 1.O L of H2O = 1.0 kg of H2O • Energy in Joules • 4.184 J x 1000 g x 75.0°C = 313,800 J g °C • = 314,000 J = 314 kJ • Energy in Calories • 314,000 J x 1 cal = 7.50 x 104 cal = 75.0 kcal • 4.184 J

  13. If 455 J of heat is transferred to 25.0 g of water at 45.0°C, what is the final temperature of the water? Let’s set of the equation, then use algebra to solve for temperature • 4.184 J x 25.0 g x (Tfinal-45.0°C) = 455 J g °C • (Tfinal –45.0°C) = 455 J 4.184 J x 25.0 g g °C • (Tfinal–45.0°C)= 455 104.6 • Tfinal = 45.0 + 4.34 = 49.3 °C

  14. Heat of Solution TheHeat of Solutionis the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent.

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