Warmup Given the following equations: H 3 BO 3(aq) HBO 2(aq) + H 2 O (l) ΔH rxn = -0.02 kJ H 2 B 4...

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Warmup Given the following equations: H 3 BO 3(aq) HBO 2(aq) + H 2 O (l) ΔH rxn = -0.02 kJ H 2 B 4 O 7(aq) + H 2 O (l) 4HBO 2(aq) ΔH rxn = - 11.3 kJ H 2 B 4 O 7(aq) 2B 2 O 3(s) + H 2 O (l) ΔH rxn = 17.5 kJ find the Δ H for this overall reaction: 2H 3 BO 3(aq) B 2 O 3(s) + 3H 2 O (l) Answer: 14.4 kJ
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Transcript of Warmup Given the following equations: H 3 BO 3(aq) HBO 2(aq) + H 2 O (l) ΔH rxn = -0.02 kJ H 2 B 4...

  • WarmupGiven the following equations:H3BO3(aq) HBO2(aq) + H2O(l) Hrxn = -0.02 kJH2B4O7(aq) + H 2O(l) 4HBO2(aq) Hrxn = -11.3 kJH2B4O7(aq) 2B2O3(s) + H2O(l) Hrxn = 17.5 kJ

    find the H for this overall reaction:2H3BO3(aq) B2O3(s) + 3H2O(l)

    Answer: 14.4 kJ

  • Specific Heat Capacity and Latent Heat

  • Every substance has its own specific heat capacity because substances respond differently to being heatedA molecule of water can absorb much more heat than an atom of metal without increasing in temperature. Water can absorb 4.180 J per gram of water. That 1g water will increase by 1C. This is its specific heat capacity(c) .

  • q = m c T

    q stands for

    c stands for

    m stands for

    T stands for heat energy (J)specific heat (J/gC)mass (g)temperature change (C)Specific heat capacity: the amount of energy required to raise the temperature of one gram of a substance by one degree celsius.

  • Ex 1: A 21 g cube of silver is heated from 30.0C to 460C. How much heat did the silver absorb? (csilver = 0.240 J / g C)q = c x m x TT = Tf - Ti

    ?0.240 J/gC 21 g 16 Cq = 0.240 J/gC x 21 g x 16 Cq = 81 Jor 0.081 kJ

    qcmT

  • Ex 2: A 21.4 g sample of gold absorbs 152.8 J of energy when its temperature is raised from 23.50C to 78.9 C. Find the specific heat of gold.q = c x m x T?152.8 J 21.4 g 55.4 Cc = 0.129 J/gC

    qcmT

  • Ex 3: If 600. Joules are applied to 250. grams of liquid water at 16.0000C, what will be the new temperature of the water? (cwater = 4.18 J/g C)q = c x m x T?600. J 250. g 4.18 J/g CT = 0.574 CTf = 16.574 C

    qcmT

  • Ex 4: The temperature of a cup of hot tea is 50.C and normal body temp is 37C. If you drink a 200.0g cup of tea, how much heat energy transfers to your body? Assume your whole body temperature increases by 0.1C heat lost by = heat gained drink by body -qd = qb-mcT = qb-(200.0g)(4.180J/gC)(-12.9C) = qb qb = 10784 J.roundYour body will absorb 11,000 J

  • Complete the diagrams below with the appropriate phase change names:endothermic phase changes require (absorb) energy from the surroundingsexothermic phase changes release (give off) energy to the surroundingsdepositionfreezingsublimationcondensationvaporizationmelting

  • What do you notice about the diagram below?During a phase change, there is NO T.All heat applied during a phase change is used to break the intermolecular forces that keep the molecules togetherLatent heat is hidden heatno change in temperature.

    water and steamice and water

  • Ex 1: How much heat must be added to a 25g ice cube at 0C to completely melt it to water at 0C?Heat of fusion: amount of energy required to completely melt 1 gram of a substance (freezing would use the same constant, but negative). q = mHf q = (25g)(334 J/g)

    8400 JImportant Information for water:Hf = 334 J/gHv= 2260 J/gCice = 2.06 J/gCCliquid = 4.18 J/gCCvapor = 1.87 J/gCYoull be given all this stuff on the test!

  • Ex 2: How much heat must be added to a 25g sample of liquid water at 100C to completely vaporize it to water into steam at 100C?Heat of vaporization: amount of energy required to completely vaporize 1 gram of a substance (condensation would use the same constant, but negative).q = mHvq = (25g)(2260 J/g)

    q = 57000 JImportant Information for water:Hf = 334 J/gHv= 2260 J/gCice = 2.06 J/gCCliquid = 4.18 J/gCCvapor = 1.87 J/gC

  • Ex 3: How much energy is lost when 20.0 g of water decreases from 303.0 C to 283.0 C?q = mcT = (20.0g)(1.87 J/gC)(-20.0C)

    q = - 748 Jor, 748J are lostImportant Information for water:Hf = 334 J/gHv= 2260 J/gCice = 2.06 J/gCCliquid = 4.18 J/gCCvapor = 1.87 J/gC

  • Ex 4: What is the total change in energy when 100.0g steam at 100.C condenses to water at 80. C? q = - mHvq = - (100.0g)(2260 J/g)q = mcTq = (100.0g)(4.18J/g C)(-20.)qtotal = -234360 J or -2.3 x 105 JImportant Information for water:Hf = 334 J/gHv= 2260 J/gCice = 2.06 J/gCCliquid = 4.18 J/gCCvapor = 1.87 J/gC= - 226000 J= - 8360 J

  • Ex 5: What is the TOTAL amount of heat necessary to bring 15.0g ice at -15.0C up to steam at 135 C?q = mcT = (15.0g)(2.06 J/gC)(15.0C) = 464 Jq = mHf = (15.0)(334 ) = 5010 Jq = mHv = (15.0)(2260) = 33900 Jq = mcT =(15.0)(1.87)(35.0) = 982 Jq = mcT =(15.0)(4.18)(100.0) = 6270 J

    qTOTAL = 46626 J4.66 x 104 J

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