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Energy Absorbed Energy Released Heat, Calorimetry, & Heating Curves GASLIQUID SOLID.

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Presentation on theme: "Energy Absorbed Energy Released Heat, Calorimetry, & Heating Curves GASLIQUID SOLID."— Presentation transcript:

1 Energy Absorbed Energy Released Heat, Calorimetry, & Heating Curves GASLIQUID SOLID

2 Heat = “transfer” of thermal energy It’s incorrect to say matter “contains” heat. heat Matter “contains” KE 80 o C20 o C Energy of Motion Stored Energy Thermal Energy Chemical Energy (moving or vibrating)(bonds & attractions) and PE.

3 The same amount of heat is added to each container, but the temperature of the container with less water increases more. Describe: WHY? Temp is higher b/c a measure of _____ KE. Explain: avg

4 + 4.18 J of heat heat required to raise 1 gram by 1 o C specific heat capacity(c): Different substances have different capacities for storing energy. The water in food holds 4x the energy as an aluminum pan. c (water) = ________ 4.18 J/g o C

5 Water absorbs/releases a lot of heat for a small ∆T. (change in temp) 1 g of water requires more heat than 1 g of metal for every 1 o C change. (higher specific heat capacity) Water has a high c ! Why?

6 Water “soaks up” (stores) a lot of energy in intramolecular movements and intermolecular attractions (H-bonds) without changing average KE (Temp) (demo) Al metal atoms vibrate in place vibratingstretching rotating water has a high c !

7 Heat transferred can be measured and calculated. q = heat (J) m = mass (g) c = specific heat (J/g o C) (4.18 J/g o C for water) ∆T = change in T (T final – T initial ) Calorimetry given on exam q = mc  T

8 Sample Calculation: How much heat is needed to warm 500 g of water from 25 o C to 100 o C? q = mc  T q = (500)(4.18)(100 – 25) q = 157,000 J c = 4.18 J/g o C

9 Ice melts by absorbing heat, but without increasing temperature. 0 o C solid  0 o C liquid (avg KE stays same) HOW? …by changing phase.

10 vaporize/condense melt/freeze same temp during phase change (=KE avg ) What happens to added heat? added heat breaks IMAFs & increases distance (↑PE) Heating Curve

11 100 o C 0oC0oC Heating Curve of Water Temp ( o C) Time (or Heat Added) gas liquid solid condensation boiling freezing (↑KE) (↑PE) endothermic exothermic melting point freezing point boiling point melting heat breaks IMAFs heat raises Temp handout heat raises Temp heat breaks IMAFs KE PE

12 heat of vaporization 100 o C 0oC0oC Heating Curve of Water Time (or Heat Added) gas liquid solid ( ∆ H fus ) heat required to melt (or freeze) heat required to vaporize (or condense) heat of fusion ( ∆ H vap ) (2260 J) (418 J) (334 J) 4.18 J/g o C Why is ∆ H vap >>> ∆ H fus ? L  G takes more energy to break IMAFs than S  L Temp ( o C) handout

13 100 o C 0oC0oC Cooling Curve of Water Time (or Heat Removed) gas liquid solid Temp ( o C) The phase change sequence is reversible. condensation boiling freezing melting

14 animation

15 1.Temperature is directly proportional to the _________ of a substance. A.thermal energy B.vibrational kinetic energy C.average kinetic energy D.total kinetic energy Quick Quiz!

16 2.Heat is simply another word for _________. A.temperature B.internal energy C.energy transferred from hot to cold D.thermal energy transferred Quick Quiz.

17 3.How much energy would it take to raise 1 gram of liquid water from 20 o C to 30 o C? A.4.18 J B.20.9 J C.41.8 J D.209 J Quick Quiz. + 4.18 J

18 4.How many joules would it take to raise 10 grams of liquid water from 20 o C to 30 o C? A.2090 J B.418 J C.209 J D.41.8 J Quick Quiz. + 4.18 J

19 5.Molecules with stronger IMAFs will have… A.higher heats of vaporization B.higher heats of fusion C.higher boiling points D.higher melting points E.higher viscosity F.higher specific heat capacity G.higher surface tension H.ALL of the above Quick Quiz.


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