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Warmup

Warmup. 689 = 32.04x. 1) Solve for x 689 = (1.8)(x )(45.2 – 27.4) 689 = (1.8)(x)(17.8)

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Warmup

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  1. Warmup 689 = 32.04x 1) Solve for x 689 = (1.8)(x)(45.2 – 27.4) 689 = (1.8)(x)(17.8) 2) On a hot summer afternoon, you get into the car, and it’s unbearably HOT inside! You grab your seat belt to buckle up and drink from a water bottle that has been sitting in the car. Think about your experiences with these two items. x = 22

  2. Specific Heat Capacity

  3. Every substance has its own specific heat capacity because substances respond differently to being heated A 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 1⁰C. This is it’s specific heat capacity(c) .

  4. Specific heat capacity: the amount of energy required to raise the temperature of one gram of a substance by one degree celsius. q = m c∆T qstands for cstands for mstands for ∆Tstands for heat energy (J) specific heat (J/g°C) mass (g) temperature change (°C)

  5. 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 ∆T ∆T = Tf - Ti q = 0.240 J/gCx 21 gx 16 °C ? 0.240 J/gC q = 81 J or 0.081 kJ 21 g 16 °C

  6. 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 °C c = 0.129J/gC

  7. 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 4.18 J/g• °C 250. g ? ∆T = 0.574C Tf= 16.574 C

  8. Ex 4: The temperature of a cup of hot tea is 50.°C and normal body temp is 37°C. If you drink a 200.0g cup of tea, how much heat energy transfers to your body? Assume your whole body temperature increases by 0.1°C heat lost by = heat gained drinkby body -qd = qb -mc∆T = qb -(200.0g)(4.180J/gC)(-12.9C) = qb qb = 10784 J….round… Your body will absorb 11,000 J

  9. Complete the diagrams below with the appropriate phase change names: exothermic phase changes release (give off) energy to the surroundings endothermic phase changes require (absorb) energy from the surroundings melting condensation deposition sublimation vaporization freezing

  10. What do you notice about the diagram below? During a phase change, there is NO ΔT. • All heat applied during a phase change is used to break the intermolecular forces that keep the molecules together • “Latent heat” is hidden heat…no change in temperature. water and steam ice and water

  11. Ex 1: How much heat must be added to a 25g ice cube at 0ºC to completely melt it to water at 0ºC? Heat of fusion: amount of energy required to completely melt 1 gram of a substance (freezing would use the same constant, but negative). q = m∆Hf q = (25g)(334 J/g) 8400 J Important Information for water: ∆Hf = 334 J/g ∆Hv= 2260 J/g Cice = 2.06 J/g∙°C Cliquid = 4.18 J/g∙°C Cvapor = 1.87 J/g∙°C

  12. Ex 2: How much heat must be added to a 25g sample of liquid water at 100ºC to completely vaporize it to water into steam at 100ºC? Heat of vaporization: amount of energy required to completely vaporize 1 gram of a substance (condensation would use the same constant, but negative). q = m∆Hv q = (25g)(2260 J/g) q = 57000 J Important Information for water: ∆Hf = 334 J/g ∆Hv= 2260 J/g Cice = 2.06 J/g∙°C Cliquid = 4.18 J/g∙°C Cvapor = 1.87 J/g∙°C

  13. Ex 3: How much energy is lost when 20.0 g of water decreases from 303.0 °C to 283.0 °C? q = mcΔT = (20.0g)(1.87 J/g°C)(-20.0°C) q = - 748 J or, 748J are lost Important Information for water: ∆Hf = 334 J/g ∆Hv= 2260 J/g Cice = 2.06 J/g∙°C Cliquid = 4.18 J/g∙°C Cvapor = 1.87 J/g∙°C

  14. Ex 4: What is the total change in energy when 100.0g steam at 100.ºC condenses to water at 80. ºC? q = - m∆Hv q = - (100.0g)(2260 J/g) q = mc∆T q = (100.0g)(4.18J/g ° C)(-20.°) qtotal = -234360 J or 2.3 x 105 J = - 226000 J = - 8360 J Important Information for water: ∆Hf = 334 J/g ∆Hv= 2260 J/g Cice = 2.06 J/g∙°C Cliquid = 4.18 J/g∙°C Cvapor = 1.87 J/g∙°C

  15. q = m∆Hv = (15.0)(2260) = 33900 J q = mcΔT =(15.0)(1.87)(35.0) = 982 J Ex 5: What is the TOTAL amount of heat necessary to bring 15.0g ice at -15.0⁰C up to steam at 135 ⁰ C? q = mcΔT = (15.0g)(2.06 J/g∙°C)(15.0⁰C) = 464 J q = m∆Hf = (15.0)(334 ) = 5010 J q =mcΔT =(15.0)(4.18)(100.0) = 6270 J qTOTAL = 46626 J 4.66 x 104 J

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