ENERGYSpecific heat and phase changes

DEFINITION OF CHEMISTRY

Reminder: At the beginning of the year we defined chemistry as the study of matter and energy.

So far: We have looked at many properties of

matter (structure of atoms, molecular shapes, physical/chemical properties, etc)

Now: We look at the properties of energy in

matter

WHAT IS ENERGY? Energy: the ability to do work or

produce heat

Two types of energy Potential Energy: energy due to the

composition or position of an object (stored energy)

Kinetic Energy: energy of motion

ENERGYChemical systems contain both kinetic and

potential energy.Potential Energy: the energy stored in

the chemical bonds of a moleculeKinetic Energy: the vibration or

movement of molecules in a substanceThis is related to the temperature of an

object

IMPORTANT CONCEPTAn important concept in science is the law

that describes energyLaw of conservation of energy: In any

chemical reaction or physical process, energy can be converted from one form to another, but it is neither created or destroyed.This concept will become more important

when we discuss chemical reactions

TOPICS IN ENERGYWe will be discussing two aspects

of energy in this unit

Specific HeatChanges in States of Matter

WHAT IS HEAT?For the next minute, discuss

in your groups what you think is the definition of heat.

DEFINITION OF HEAT Heat: energy that is in the process of flowing

from a warmer object to a cooler object Therefore, heat is a relative term

95° is cool when it is summer in Vegas50° is warm when it is winter in Wisconsin

Heat is different from energy and temperature. Again, temperature is a measure of the movement of molecules

UNIT FOR HEAT AND ENERGYThe unit for both heat and energy

are the same. We will be discussing 2 units:

Calorie (cal)On your food lables, the calorie

actually represents 1000 cal or 1 kilocalorie (1 kcal)

Joule (J)

CALORIECalorie (cal): defined as the

amount of energy needed to raise the temperature of one gram of water 1°C

Joule (J): this is the SI (metric) unit of energy1 cal = 4.184J

DIMENSIONAL ANALYSISAgain, we must use dimensional

analysis to convert between units

REMEMBER: 1 cal = 4.184JThis is your conversionAlways write down the unit you

start with and cancel out the units

EXAMPLEYou release 250 calories of

energy from a chemical sample. How much energy is this in the unit of Joules?

ANSWER250 cal | 4.184 J = 1046J

1 cal

REMINDER: Sig figs.You start with 2 sig figs. (250)You need 2 sig figs. in your answer

Therefore: 1.0x103 J

TRY THESE1. A food item says it contains 2.35x103J

of energy. What is this in calories (cal)?2. You start with 399 cal of energy. What

is this in Joules (J)?3. A yogurt contains 170 calories

according to the label. What is this in Joules (J)? (REMINDER: Food label calories are really kilocalories)

WHAT ARE SOME USES OF ENERGY?

Have you ever noticed that when you boil water, the pot heats up much faster than the water in the pot?

Why does the pot heat up faster than the water?

Discuss in your groups for 1 minute why you think this is true.

SPECIFIC HEATThe reason is that water needs to

absorb more heat than the pot to increase in temperature

This property of matter is called specific heat

SPECIFIC HEAT: the amount of heat required to raise the temperature of one gram of a substance by 1°C

EXAMPLEThe specific heat of water is:

4.184 J/(g • °C)We’ll discuss the unit in a minute

The specific heat of concrete is 0.84 J/(g • °C)

Therefore the specific heat of water is about 5 times bigger than concrete

What does this mean?

ANSWEROn a hot day, concrete requires less

energy to heatTherefore, concrete heats up much

faster than waterThis is why you don’t walk on concrete

barefoot, but you can still go in the water to cool offWater doesn’t heat up very fast so it

stays cooler

HOW HOT DOES SOMETHING GET?

We defined earlier what specific heat was

We also showed the units involved in specific heat:

J/(g • °C)We will now discuss how to calculate

how hot something gets when exposed to a certain amount of heat or energy

EQUATION FOR HEATTo explain temperature change, we

have to explain the equation for calculating heat:

q = c x m x δT

DEFINING THE TERMSq: the heat absorbed or released

from a substance in Joules (J) or calories (cal)

c: the specific heat of the substancem: the mass of the substance in

grams (g)δT: the change in temperature in °C

(or Tfinal – Tinitial)

EXAMPLEAluminum has a specific heat of

0.897 J/(g • °C). If you add 250 cal of energy to 150g of aluminum, how much does the temperature increase?

STEP 1Write down your variables and make sure

they are in the proper units:q = 250 cal c = 0.897 J/(g • °C)m = 150g δT = ?NOTE: Heat is in calories and specific heat is

in Joules, we need to convert

STEP 1q = 250 cal

250 cal | 4.184J = 1.0x103 J 1 cal

c = 0.897 J/(g • °C)m = 150g δT = ?

STEP 2Plug your values into the formula:

q = c x m x δT

1.0x103J = 0.897 J/(g • °C) x 150g x δT

STEP 3Perform your calculations as you cancel

out units:1.0x103J = 0.897 J/(g • °C) x 150g x δT1.0x103J = 135 J/°C x δT

135 J/°C 135 J/°C

7.4 °C = δT

TRY THESE1. Water has a specific heat of

4.184 J/(g • °C). If you add 125J of energy to 22.2g of water, how much does the temperature increase?

2. Granite has a specific heat of 0.803 J/(g • °C). If you add 125J of energy to 22.2g of granite, how much does the temperature increase?

ANSWER

1.1.35°C2.7.01°C Therefore granite heats up about

5times faster than water

OTHER CALCULATIONS WITH SPECIFIC HEAT

In addition for solving for temperature change, you can solve for specific heat, amount of heat or the mass of the object.

Let’s explore some examples

EXAMPLE 1An object absorbs 1.33x104J of

energy. If it has a mass of 15.7g and the temperature increased 25°C, what is the specific heat of the object?

REMINDER: Make sure you have the correct units in the answer.

ANSWER 1

34 J/g•°C

EXAMPLE 2An object has a mass of 35.0g and

the temperature increased 15°C. If the object has a specific heat of 0.647 J/g•°C, how much heat was absorbed?

ANSWER 2

340 J

EXAMPLE 3A piece of lead has an initial

temperature of 25°C. The specific heat of lead is 0.129 J/g•°C. If 55.5g of the lead absorbs 250J of energy, what is the final temperature?

REMINDER: δT = (Tfinal – Tinitial)

ANSWER 3δT = 35°CTinitial = 25°C

δT = Tfinal – Tinitial

35°C = Tfinal - 25°C

Tfinal = 60°C

EXAMPLE 4An object has a mass of 12.5g and

the temperature DECREASED 5.7°C. If the object has a specific heat of 0.647 J/g•°C, how much heat was released?

NOTE: If temperature decreases, you will have a negative “-” δT

ANSWER 4

-46J

HOW DO WE MEASURE HEAT?

So far we have had a lot of example problems, solving for heat (q), change of temperature (δT), mass (m) and specific heat (c)

What method do scientists have to measure “heat”?

CALORIMETERScientists measure heat using a

calorimeter.

Calorimeter: an insulated device that is used to measure the amount of heat released or absorbed during a physical or chemical process

SUMMARY1. Almost all of the energy from the reaction vessel

is transferred to the water2. The temperature change of the water is observed

(no energy is lost because it is insulated)3. You calculate the energy gained by the water

(you know the specific heat of water, the mass of water, and can read the δT)

4. This is the amount of heat from the reaction in the chamber

EXAMPLE You heat up 3.243kg of an unknown

metal and place it in the water of a calorimeter. As the metal cools by 8.2°C, it raises the temperature of the water 15°C and then remains constant. What is the specific heat of the metal if the metal was placed in 100g of water?

KNOWN: Specific heat of water: 4.184 J/g•°C

STEP 1 Calculate the amount of heat absorbed

by the water q = c x m x δT (for water) q = (4.184 J/g•°C)(100g)(15°C) qwater = 6276J

qwater = qmetal

Therefore, qmetal = 6276J

STEP 2 Calculate the specific heat of the metal

q = 6276J c = ? m = 3.243kg = 3243g δT = 15°C

q = c x m x δT (for water) 6276J = c (3243g)(8.2°C) c = 0.236 J/g•°C

STEP 3Use the specific heat to figure out

the unknown metalLook on page 520 of the text at

your deskWhich metal was put in the

calorimeter?

ANSWER

SILVER

TRY THE FOLLOWING You heat up 1.617kg of an unknown

metal and place it in the water of a calorimeter. As the metal cools 5.5°C, it raises the temperature of the water 5.5°C and then remains constant. What is the metal, if the metal was placed in 250g of water?

KNOWN: Specific heat of water: 4.184 J/g•°C

ANSWER

CALCIUM

REVIEWEarlier in the year, we discussed

phase change diagrams

In your groups, create a phase change diagram:Solid Liquid; Liquid Gases

PHASE CHANGE DIAGRAMStays in gaseous

state. Gas gets hotter

Substance is going from liquid to gas. The

bonds holding the liquid together break

apart

Liquid is heating up. Liquid molecules

speed up

Substance is going from a solid to liquid. The bonds

holding the solid molecules together break

apart

Solid is heating up. Solid increases in

temperature

SPECIFIC HEATWe have already discussed the energy needs

for one aspect of the phase change diagramThe energy needed to increase the

temperature of a substance is dependent on the specific heat

SPECIAL NOTE: the specific heat of a substance depending on what state of matter it is in

EXAMPLE Specific heats for water:

Liquid: 4.184 J/g•°CSolid: 2.03 J/g•°CGas: 2.01 J/g•°C

CALCULATING HEAT NEEDS

Knowing the specific heats of substances allows us to determine how much energy is needed to change the temperature of a substance

What happens to energy needs when there is no change in temperature?EXAMPLE: Phase changes

SOLID LIQUIDBefore we describe the energy needs for going from solid to liquid, we need to describe how solids and liquids behave at the molecular level

SOLID LIQUID

WHAT WE KNOW ABOUT SOLIDS: Solids are close together They are not free moving (bound tightly

together) The bonds vibrate, but are fixed Solids tend to be more dense than liquid

(this is why solids tend to sink in the liquid of the same matter) EXCEPTION: Water

SOLID LIQUID WHAT WE KNOW ABOUT LIQUIDS:

Liquids are still close together They are free moving The bonds that held them together as a

solid have been broken apart Liquids are free to move around, but

cannot expand further because of intermolecular forces (δ+ and δ-) EXAMPLE: Water Hydrogen bonds

SOLID LIQUID

Therefore, to go from a solid to a liquid, you need enough energy to break the bonds holding the solid together

This is called the heat of fusionHeat of Fusion: the energy

needed to melt one gram of a solid substance

SOLID LIQUIDSince there is no temperature

change for heat of fusionAll the energy is being used to

break the bonds holding the solid together

The formula is:q = m x Hfus

EXAMPLEYou have 250g of ice. The heat of

fusion for water is 334 J/g. How much energy is need to melt the ice?q = ?m = 250gHfus = 334 J/g

q = (250g)(334 J/g) = 8.4x104 J

TRY THESE1. You have 125g of solid acetic

acid. The heat of fusion is 195 J/g. What is the heat needed to melt the acetic acid?

2. You have 0.00344g of solid ethanol. The heat of fusion is 107 J/g. What is the heat needed to melt the ethanol?

ANSWERS

1.2.44X104 J2.0.368J

LIQUID GASWHAT WE KNOW ABOUT GASES:

Gases are very far apartThey are free moving and rapid The bonds that held them together

as a liquid have been broken apartGases do not have any force of

attraction holding them together

LIQUID GAS Therefore, to go from a liquid to a gas,

you need enough energy to break apart the intermolecular forces holding the liquid together

This is called the heat of vaporization Heat of Vaporization: the energy

needed to vaporize (or boil) one gram of a liquid substance

LIQUID GASSince there is no temperature

change for heat of vaporizationAll the energy is being used to

break the intermolecular forces holding the liquid together

The formula is:q = m x Hvap

EXAMPLE You have 250g of liquid water. The heat

of vaporization for water is 2262 J/g. How much energy is need to boil the water? q = ? m = 250g Hvap = 2262 J/g

q = (250g)(2262 J/g) = 5.7x105 J

TRY THESE1. You have 125g of liquid acetic

acid. The heat of vaporization is 390 J/g. What is the heat needed to boil the acetic acid?

2. You have 0.00344g of liquid ethanol. The heat of vaporization is 836 J/g. What is the heat needed to boil the ethanol?

ANSWERS

1.4.88X104 J2.28.8 J

PUTTING IT ALL TOGETHER

Now it is possible to determine the total energy used in a system

The things to remember There is a different specific heat for each

state of matter (solid, liquid, gas) Each substance has a heat of fusion for

melting or freezing Each substance has a heat of vaporization

for boiling or condensing

EXAMPLE You have 3500g of solid water (ice) at -

15°C. How much energy is needed to convert all of this water to a gas with a temperature of 110°C? Hfus = 334 J/g Hvap = 2262 J/g csolid = 2.03 J/g•°C cliquid = 4.184 J/g•°C cgas = 2.01 J/g•°C

STEP 1Calculate the heat for increasing the

temperature of the solidMake sure to use the correct specific

heatq1 = c x m x δT

q1 = (2.03 J/g•°C) x (3500g) x (15°C)

q1 = 1.1x105 J

STEP 2Calculate the heat needed to

convert from a solid to a liquidq2 = m x Hfus

q2 = (3500g) x (334 J/g)

q2 = 1.2x106 J

STEP 3Calculate the heat for increasing

the temperature of the liquidMake sure to use the correct

specific heatq3 = c x m x δTq3 = (4.184 J/g•°C) x (3500g) x

(100°C)q3 = 1.5x106 J

STEP 4Calculate the heat needed to

convert from a solid to a liquidq4 = m x Hvap

q4 = (3500g) x (2262 J/g)

q4 = 7.9x106 J

STEP 5Calculate the heat for increasing the

temperature of the gasMake sure to use the correct specific

heatq5 = c x m x δT

q5 = (2.01 J/g•°C) x (3500g) x (10°C)

q5 = 7.0x104 J

STEP 6Add all of the heats together to

get a totalq = q1 + q2 + q3 + q4 + q5

q = 1.1x105J + 1.2x106J + 1.5x106J + 7.9x106J + 7.0x104J

q = 1.1 x107 J

SPECIAL NOTE You will not always have to do all the steps Only make the conversions you need

Example: Taking 20g of water from 105°C to 75°C

You will only have to do the following:Water gas from 105°C 100°CPhase change from gas liquidWater liquid from 100°C 75°C

TRY THE FOLLOWINGCopper Melting point:

1084°C Boiling point: 2562°C Hvap: 4688 J/g

Hfus: 202 J/g

Csolid: 0.385 J/g•°C

Cliquid: 3.45 J/g•°C

Cgas: 24.47 J/g•°C

To the left is the information for copper. If you have 1.25g of copper that starts at 1000°C, what is the energy needed to get it to 2500°C?

ANSWER FIRST 2 STEPS:

Heat up the solid:q1 = c x m x δT q1 = (0.385 J/g•°C)(1.25g)(1084°C-1000°C)q1 = 40.4 J

Convert from solid to liquidq2 = m x Hfus

q2 = (1.25g)(202J/g)q2 = 253 J

ANSWER – NOTE WE DID NOT NEED ALL STEPS

LAST 2 STEPS Heat up the liquid

q3 = c x m x δT

q3 = (3.45 J/g•°C)(1.25g)(2500°C-1084°C)

q3 = 6.11x103 J

Add all the heats q = q1 + q2 + q3

q = 40.4J + 253J + 6.11x103 J q = 6.40x103 J