Conservation of Energy Like mass, energy cannot be created or destroyed, just transferred from one place to another within the system The most common energy transfer is in the form of heat Transfer of heat is measured as a temperature change Temperature is a measure of the average kinetic energy of the particles in the substance

Measuring Heat SI unit for heat is Joules Specific heat 4.184 Joules = 1 calorie Calorie is the amount of heat it takes to raise 1 gram of water by 1 degree Celsius Specific heat The amount of heat necessary to raise one gram of a substance by one degree Celsius Unique for each substance The specific heat of water is 1 calorie/gºC Also 4.184 J/ gºC

Kinetic Theory As the energy of the particles in a substance increases Particles move faster Density decreases As the energy of the particles in a substance decreases Particles move more slowly Density increases

Calculating Heat Heat is calculated using the heat equation Q = (m) (ΔT) (cp) Q = heat M = mass ΔT = change in temperature Tf – Ti (final temp – initial temp) Cp = specific heat

Example Problem How much heat is lost when a solid aluminum ingot with a mass of 4110 g cools from 660.0 C to 25.0 C? q=(m) (T) (Cp) Mass is 4110g Initial temp is 660.0 C Final temp is 25.0 C Specific heat of aluminum is 0.903 J/gC Q = (4110) ( 660C – 25.0 C ) ( 0.903 J/gC) Q = (4110) ( 635C) ( 0.903 J/gC) Q = 2.35 x 106

Energy and Change of State If the change in kinetic energy is large enough this will result in a change of state As the amount of energy changes, the temperature will increase or decrease until the boiling (heat increase) or freezing (heat loss) point is reached At the freezing or boiling point two phases of matter can exist at the same temperature

To make the change from one phase to another more energy will be absorbed (boiling or melting) or lost (condensing or freezing) without a corresponding change in temperature This is because this energy is used merely to overcome the intermolecular forces maintaining one state and move to the new state

Heat of vaporization Hv Heat necessary to change from liquid to gas or from gas to liquid at the boiling/condensation point Measurement of the amount of potential energy absorbed /lost when a change of state occurs between liquid and gas

Heat of fusion Hf Heat necessary to change from liquid to solid or from solid to liquid at the freezing/melting point Measurement of the amount of potential energy absorbed /lost when a change of state occurs between liquid and solid

Phase change graph

Slanted sections of the graph represent change in kinetic energy M ΔTCp Flat portions represent the change in potential energy necessary to change from one phase or state of matter to another – Hf or Hv; look at it again

Phase change graph

If one were to begin with a cold solid, and add heat, each step as heat is added must be calculated The solid warms up to the melting point M ΔTCp (using the specific heat for the solid) At the melting point, some liquid will begin to form the heat necessary to melt it completely is calculated by mass of the solid times the heat of fusion

The liquid increases temperature until reaching the boiling point calculated by M ΔTCp (using the specific heat for the liquid) At the boiling point gas will begin to form the heat necessary to vaporize it completely is calculated by mass of the solid times the heat of vaporization

Vapor may continue to increase in temperature calculated by M ΔTCp (using the specific heat for the gas) The sum of all of these heats (from initial temperature to final temperature) is used to calculate the heat gained or lost by the substance

If a 26. 5g block of ice at –15. 0 ºC is heated to water at 20 If a 26.5g block of ice at –15.0 ºC is heated to water at 20.0 ºC, how much heat is used? Ist heat the ice to its melting point M ΔTCp (26.5g)(0.0C – (-15.0C))(0.53cal/gºC) 210.7 cal

Now melt the ice Now warm the water M(Hf) 26.5g (80 cal/g) 2120 cal M ΔTCp 26.5g (20 ºC - 0 ºC)(1.00cal/g ºC) 530 cal

Now add up the calories 210.7 cal + 2120 cal + 530 cal 2860.7 cal 2860 cal (2 sig fig.) Now try one yourself

How much heat is needed to heat 117g of water at 42 ºC to steam at 136 ºC

117g (100 ºC - 42 ºC)(1.00 cal/g ºC) 6786 cal 117g (540 cal/g) 63180 cal 117(136 ºC - 100 ºC)(0.480 cal/g ºC) 2021.76 cal 6786 cal + 63180 cal + 2021.76 cal 71988 cal 7.20 x 104 cal

How much heat does it take to raise the temp of 112g of ice at –4 How much heat does it take to raise the temp of 112g of ice at –4.0 ºC to 120 ºC?