Do Now:

Why is CO2 a gas and H2O a liquid at room temperature?

What can we do to change H2O to a gas?

How can we change CO2 to a liquid?

Heat and the States of Matter

Homework:Read pages293 – 299

Answer the following questions on page 299 #4, 6, 7, 9

ANDThe even # questions on the handout

Heat

• Def: the total amount of kinetic energy of the particles in a sample of matter

• Units: Joule (J)

• Heat always flows from objects at higher temps to objects at lower temps, until both objects are at the same temperature

Temperature

• Def: a measure of the average kinetic energy of the particles in a sample of matter (not a form of ENERGY)

Measuring Heat

• Calorimeter is the device used to measure the quantity of heat lost or gained in a reaction– Endothermic – heat

absorbed– Exothermic – heat

released

Specific Heat Capacity (c)

Def: the amount of heat needed to raise the temp of 1 gram of a substance 1 ⁰C

- All substances have different specific heat capacities depending on their composition

- Water’s specific heat capacity is listed on Table B

Calculating heat lost or gained

q = m c D t

Where, q = heat lost or gained (J)c = specific heat capacity (J/g⁰C)D t = change in temperature (⁰C)(final temp - initial temp)

How much heat is required to raise 50.0 g of water from 5.00 ⁰C to 20.0 ⁰C?

q =

m =

c =

ti =

tf =

Q = mcDt

= (50.0 g)(4.18J/g C) (20.0 C – 5.00 C)

= 3138 J

= 3140 J

How much heat is required to raise the temp of 200.0g of water 24.5C?

If 500.0 J of heat are added to 10.0 g of water at 30.0 C. What is the

final temp of the water?

Using a Calorimeter

• While studying heat exchanges we define the experiment that is being performed as the system, and everything outside of the experiment as the surroundings.

Law of Conservation of Energy

• Energy can neither be created nor destroyed in a chemical reaction but may be converted from one form to another.

Heat lost = Heat gained

The picture shows a reaction to be carried out in a calorimeter.

The heat released by the reactants (system) is absorbed by the surrounding water.

Measuring the temperature change of the water will allow us to calculate the heat lost during the reaction.

We can calculate the heat loss using q = mcDt for the water.

The following is a description of an experiment used to determine the specific heat of copper.

• In a laboratory investigation, a 50.0-gram sample of copper is at 100.0°C in a boiling water bath.

• A Styrofoam cup with a lid is used as a calorimeter.

• The cup contains 100.0 grams of distilled water at 23.2°C.

• The hot copper is poured into the cup of water, and the cup is quickly covered with the lid.

• A thermometer is inserted through the lid.

• The coppe rand water are gently stirred in the cup.

• The temperature is checked periodically.

• The highest temperature noted is 26.3°C.

• In terms of energy flow, explain why the temperature of the water in the calorimeter increases.

• Using the information given, complete the data table provided in your answer booklet.

Calculate the number of joules of heat gained by the water.

Calculate the specific heat capacity of the copper.

Homework

Go to Chemthink and complete the

Particulate Nature of Matter section under the introduction section.

Complete both the tutorial and the

question set

Phases of Matter

The Kinetic Molecular Theory• Matter is composed of smaller particles

– Physical properties determined by distance between particles, and forces (IMF) they exert on one another

• Particles are in constant motion– Average kinetic energy is determined by temperature

• Collisions between particles transfer energy without a loss of energy

• As substances change from a solid liquid gas, the amount of motion of the particles increases

• As substances change from solid liquid gas, the particles have enough energy to slowly overcome the forces of attraction between particles

Solids

• have a definite shape, definite volume (incompressible)• particles of substance are closely packed together• particles vibrate• 2 types of solids– Crystals – particles of solid arranged in a repeating

geometric patternex) NaCl, Diamond

– Amorphous solids – particles are not arranged in any order

ex) wax, glass

Liquids

• definite volume, no definite shape (assume the shape of container, relatively incompressible)• particles of a substance not held together

tightly• particles are in constant motion (can slide

past one another)

GasNo definite shape, no definite volume (expand and

contract, compressible)particles are widely separatedGases exert a pressure by striking the walls of a container

(each collision exerts a force on the container)Measured using a barometerAir pressure measured torr the air pressure at sea level is

760 torr, 1 atmosphere, or 101.325 kPaVolume of a gas can change under different conditions of

temp and pressure Standard temp and press (STP) have been assigned

OC and 101.3 kPa (1 atm)

Phase Changes

Solid / Liquid • Freezing – the physical change of state from a liquid to a

solid- energy is released

liquid solid + heat energy

• normal freezing point – the temperature at which a liquid changes to a solid at 101.3 kPa (normal atmospheric pressure)

• temperature does not change during phase change

Solid/Liquid phase changes

• Melting – the change of state from a solid to a liquid- energy is added to solid, giving the particles sufficient energy to break free from other particles, no temperature change

solid + heat energy liquid

• occurs at the same temp as normal freezing point

Heat changes during melting/freezing

• Heat of Fusion (Hfus) – the amount of heat energy needed to change a mass of solid into a certain mass of liquid at constant temp

for H2O (s) Hfus = 334 J/g (table B)

• when a mass of H2O (l) freezes, 334 J/g are released

• when a mass of H2O(s) melts, 334 J/g are absorbed

• to find the amount of heat energy lost or gained during melting or freezing use:

q = mass x Hfus

If 32 g of ice are melted at 0 C, how much heat has been absorbed by the ice?

q = m x Hfus

q = 32 g x 334 J/g

q = 10688 J

q = 11000 J

How many grams of ice at 0C can be melted to water at 0C by absorbing 3347 J

of heat?q = m x Hfus

q = mHfus

3347 J = m334 J/g

10.0 g = m

Why does a balloon stay inflated?

(explain in terms of the behavior of the molecules that fill the balloon)

Liquid/Gas

• Vaporization – change from a liquid to a gas– Boiling - change of a liquid to bubbles of vapor that

appear throughout

- temp at which a liquid boils is known as normal boiling point– Evaporation – vaporization that takes place at the surface

of a liquid- occurs at any temp

- particles at surface with greater average kinetic energy overcome the intermolecular forces of attraction that keep particles in liquid state, and escape into gaseous state

Liquid/Gas

• Condensation – reverse of vaporization

Gas liquid + heat

Heat calculations for boiling/condensation

Heat of vaporization – amount of heat required to change a mass of a liquid to a mass of gas at constant temp

* for water Hvap = 2260 J/g (Table B)

at 100C

q = mass x Hvap

How many joules of heat are needed to change 20. g of H2O at 100C to steam at 100C?

q = m x Hvap

q = 20. g x 2260 J/g

q = 45000 J

But wait! There’s more

Vapor Pressure – the pressure exerted by a gas from the liquid, when confined above the liquid at a given temperature

• as temp increases, KE increases, • more particles in gaseous state, • vapor pressure increases• Table H

Vapor pressure and boiling points

• ****Boiling point – the temperature at which the vapor pressure of a liquid is equal to the atmospheric pressure

• at high altitudes, atmospheric pressure is lower than at sea level

the boiling point is lower

ex) Salt Lake City and Denver, the BP of water is 95C

Heating Curve

Cooling curve

Sublimation

Phase change from a solid to a gasOccurs in substances with weak

intermolecular forces of attractions and high vapor pressures

Ex) dry iceCO2(s) + energy → CO2(g)

Deposition is the opposite process

What would the heating curve look like for a substance that undergoes sublimation?