1. Energy

2. Energy – the ability to do work Energy – the ability to do work Kinetic – energy of motion, anything that moves has kinetic energy. Kinetic – energy of motion, anything that moves has kinetic energy. K.E. = 1/2mv 2 K.E. = 1/2mv 2 Potential – is stored energy, chemical or mechanical Potential – is stored energy, chemical or mechanical

3. Law of Conservation of energy or law of conservation of matter and energy Law of Conservation of energy or law of conservation of matter and energy Matter and energy can not be created or destroyed; it can only be converted from Matter and energy can not be created or destroyed; it can only be converted from One form to another. E = mc 2 One form to another. E = mc 2

4. Heat (q) is a measure of energy, in the metric system the unit of measure is a calorie. A calorie is the amount of heat required to raise the temperature of one gram of water one degree Celsius. In the SI system the unit is a joule. 1 calorie = 4.184J So heat is a measure of the amount energy while temperature is more of an Indication of the direction of flow of the energy. It always flows from the object With the higher temperature to the lower one.

5. Specific heat Specific Heat – the amount of energy needed to raise one gram of the material one degree Celsius. For water its 4.184j/(g.°C) or 1c/(g.°C) Specific Heat – the amount of energy needed to raise one gram of the material one degree Celsius. For water its 4.184j/(g.°C) or 1c/(g.°C)

6. Calculating heat absorbed Calculating heat absorbed q = c x m x ∆T some times ∆H is used in place of q q = c x m x ∆T some times ∆H is used in place of q q is the heat change, c is the specific heat, m is the mass of the sample and q is the heat change, c is the specific heat, m is the mass of the sample and ∆T is the change in temperature of the sample. ∆T is the change in temperature of the sample.

7. Example Example How much energy is required to warm 250.grams of water from 20.0°C to How much energy is required to warm 250.grams of water from 20.0°C to 100.°C for a cup of coffee? Q = 4.184j/g.°C x 250.grams x 80.0°C Q = 4.184j/g.°C x 250.grams x 80.0°C Q = 83680 → 83700joules (to 3 sig figs )

8. You should be able to solve for any of the variables in the equation given the other three values or a way to find them. You should be able to solve for any of the variables in the equation given the other three values or a way to find them. C = q / m x ΔT m = q / c x ΔT ΔT = q / c x m C = q / m x ΔT m = q / c x ΔT ΔT = q / c x m Q = c x m x ∆T Q = c x m x ∆T m x ∆T m x ∆T m x ∆T m x ∆T

9. Calorimetry - a method of determining the heat change of a chemical or physicl change. Calorimeter – well insulated container used in calorimetry.

10. Reactions can be endothermic, absorb energy or exothermic which release energy. (Hypothermia )

11. Phase change graph H f = 334j/g Hvap = 2260j/g sph ice = 2.03j/g H f = 334j/g Hvap = 2260j/g sph ice = 2.03j/g sph liq = 4.184j/g sph steam = 2.01j/g sph liq = 4.184j/g sph steam = 2.01j/g Phase change graph and energies