energy and energy transformations. first law of thermodynamics energy is never created nor...
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THERMODYNAMICS
Energy and energy transformations
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First Law of Thermodynamics Energy is never created nor destroyed
Energy can change forms, but the quantity is always constant.
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Second Law of Thermodynamics
The Entropy of the Universe is always increasingEntropy= disorder
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Third Law of Thermodynamics
The entropy of an ideal solid at zero Kelvin is zeroAll molecular motion stops at 0 K
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Endothermic Reactions Energy is used to
begin a reaction Products have
higher energy than reactants
Absorbs heat from surroundings Ice meltingWater evaporating
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Exothermic Reactions Gives off energy
during a reaction Reactants have
more energy than products
Gives off heat Ice freezingWater condensing
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Heat vs Temperature
Temperature - measure of average KE
Heat - measure of energy transfer
Temp change (∆T) depends on: amount of heat
transferred (q) mass of object (m) specific heat of the object
(C) Video link: http://
www.youtube.com/watch?v=wTi3Hn09OBs&safety_mode=true&persist_safety_mode=1
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Measuring Heat q = m C ∆T
q = heat○ measured in joules (J)
or kilojoules (kJ) ○ 1000J = 1kJ
m = mass; measured in grams (g)
∆ T = change in temp○ measured in Celsius
(°C) or Kelvin (K)C = specific heat
○ units are J/g°C or J/gK
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Endo/exothermic -q = release heat (exothermic) +q= absorb heat (endothermic)
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What amount of heat is needed to increase the temp of 10 g of Hg by 5C? (specific heat of Hg is 0.139 J/gC)
If 68,000J of heat are added to 25g of H2O (specific heat of water is 4.184 J/gC), what is the change in temperature in Celsius?
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Calorimetry Calorimetry
uses a closed system (calorimeter) to determine the energy change or specific heat of an unknown substance
Calorimeter- an insulated device usually filled with water or a substance with a known specific heat
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1st law - conservation of energy When applied to a closed system, any
energy that is lost by one substance is gained by the other
qsubstance1 = - qsubstance2
since q = m C ∆ T…
m1 C1 ∆ T1 = - m2 C2 ∆T2
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Ex: A 55.8g piece of unknown metal at 180°C is placed into 100.0g of water that began at 25°C. The final temperature of both was 26.8°C. If the specific heat of water is 4.184 J/g°C, calculate the specific heat of the metal.