#  Energy is transferred ◦ Exothermic – heat is released  Heat exits ◦ Endothermic – heat is required or absorbed by reaction  Measure energy in Joules.

## Presentation on theme: " Energy is transferred ◦ Exothermic – heat is released  Heat exits ◦ Endothermic – heat is required or absorbed by reaction  Measure energy in Joules."— Presentation transcript:

 Energy is transferred ◦ Exothermic – heat is released  Heat exits ◦ Endothermic – heat is required or absorbed by reaction  Measure energy in Joules or calories

 Temperature – average energy of molecular motion ◦ Size and type of particle do not matter  Heat – total energy of molecular motion ◦ Speed, amount and type of particle matter

 The heat needed to raise the temperature of 1g of a substance by 1 o C  Water has a high heat capacity of 4.184 Joules or 1 calorie

Q = heat (gained or lost) m = mass (in grams) c = specific heat value ∆T = change in temperature

 How much heat is lost when solid aluminum ingot with a mass of 4110g cools from 660 o C to 25 o C? The specific heat of aluminum is 0.903 J/(g x o C).

m = 4110g c = 0.903 J/(g x o C) ∆T = 660 o C – 25 o C = 635 o C Q = ? Q = 2, 356, 695 J Or... 2, 357 kJ

 What happens to the change in temperature or equal masses of copper and water when equal amounts of heat energy are given?  c for Cu = 0.387 J/(g x o C)  Use mass = 1.0g Q=10.0J Cu H2OH2O ∆T = 25.8 o C ∆T = 2.4 o C

 Calorimeter – a device used to measure the energy given off or absorbed during a chemical or physical change.

 A piece of unknown metal with mass 23.8 g is heated to 100.0 o C and dropped into 50mL of water at 24.0 o C. The final temperature of the system is 32.5 o C. What is the specific heat of the metal? The density of water is 1g/mL.

MetalWater m = 23.8gm = 50.0g ∆T = 100 – 32.5 = 67.5 o C∆T = 32.5 – 24 = 8.5 o C c = 4.18 J/(g x o C) For water D = 1g/mL X = 50.0g Heat gained by water = Heat lost by metal c = 1.1 J/(g x o C)

 The amount of energy needed to convert 1g of a substance from liquid to gas or from gas to liquid.

 How much heat is required to vaporize 15g of liquid water?  H v of water = 2260 J/g Q = 33,900J or 33.9 kJ

 The amount of energy needed to melt 1g of solid substance to liquid.  The amount of energy released when 1g of a liquid freezes or becomes solid.

 How much energy is released when 25g of liquid water freezes?  H f of water = 334 J/g Q = 8,350 J or 8.35 kJ

 A 30g sample of water is heated from 75 o C to 135 o C. How much energy is needed?  Remember, you are crossing a phase change so you must account for that!  Use both specific heat and heat of vaporization.

This accounts for liquid portion of water. Q = 3,135 J This accounts for the phase change (liquid to gas) Q = 67,800 J Q = 2,121 J Accounts for gas (steam) portion of water.

3,135J + 67,800J + 2,121J = So, total energy is: 73,056J or 73 kJ

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