Conservation of EnergyLike 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

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 TheoryAs 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 HeatHeat 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/gC Q = (4110) ( 660C – 25.0 C ) ( 0.903 J/gC) Q = (4110) ( 635C) ( 0.903 J/gC) 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 stateAs 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 20.0 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 M(Hf) 26.5g (80 cal/g) 2120 cal

Now warm the water 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 –4.0 ºC to 120 ºC?