Energy Unit 8, Chapter 10. Energy, Temperature, and Heat Section 1.

Energy

Unit 8, Chapter 10

Energy, Temperature, and

Heat

Section 1

Energy

• Energy: the ability to do work or produce heat.

• Work: force acting over a distance.

• 2 major types:

• Potential energy: energy due to position or composition.

• Kinetic energy: energy due to motion.

The Law of Conservation of Energy

Energy can be converted from one form to another but can be neither created nor

destroyed.

Temperature, Heat, and Thermal Energy

• Temperature: a measure of the random motions (average kinetic energy) of the components of a substance.

• H2O molecules in warm water move faster than H2O molecules in cold water.

Classzone Animation


• In the animation, energy from the faster moving particles is transferred through the wall to the slower moving particles.

• The two samples will eventually reach the same temperature.

• The amount of energy lost by the hot water must be equal to the amount of energy gained by the cold water. (law of conservation of energy)


• Heat: a flow of energy due to a temperature difference; a transfer of thermal energy.

• Thermal Energy: energy due to the random motions of the components of an object.

• Most chemical reactions involve changes in energy.

• In order to study this, the universe is divided into 2 parts: the system and the surroundings.

• The system is the part of the universe on which we wish to focus attention.

• The surroundings include everything else in the universe.

SURROUNDINGS

SYSTEM

• Exothermic – refers to a process in which energy (as heat) flows out of the system into the surroundings.

- “exo” = out of

- the system releases heat

• Endothermic – refers to a process in which energy (as heat) flows from the surroundings into the system.

- “endo” = into

- the system absorbs heat

2H2 (g) + O2 (g) 2H2O (l) + energy

Endothermic

H2O (g) H2O (l) + energy

Exothermic

energy + 2HgO (s) 2Hg (l) + O2 (g)

energy + H2O (s) H2O (l)

The energy gained by the surroundings in an exothermic reaction must be equal to the

energy lost by the system.(Law of Conservation of Energy)

In any exothermic reaction, some of the potential energy stored in the chemical bonds is converted to thermal energy

(random kinetic energy) via heat.

Exothermic or Endothermic?(Remember to think about the direction!)

1. Your hand gets cold when you touch ice.exothermic

2. Ice melts when you touch it.endothermic

3. Your hand gets colder when you touch ice.exothermic

4. Propane burns in a propane torch.exothermic

5. A beaker feels cold when two chemicals are mixed in it.

endothermic

The Flow of Energy

Section 2

Thermodynamics

• Thermodynamics is the study of energy.

• The first law of thermodynamics states that the energy of the universe is constant.

• Energy is neither created nor destroyed.

• Energy can flow into (or out of) a system, but the amount gained (or lost) by the system is equal to the amount lost (or gained) by the surroundings.

Thermodynamics

• Energy is a state function, or a property of a system that does not depend on the pathway

• Example: elevation vs. distance traveled

• Application: ball down a hill

• Work and heat are not state functions

Internal Energy (E)

• The internal energy of a system is the sum of the kinetic and potential energies of all particles in the system.

• The internal energy of a system can be changed by a flow of work, heat, or both.

• Thus,E = q + w

the energy change of the system is equal to the heat plus the work

Thermodynamic Quantities

• Thermodynamic quantities are made up of 2 parts:

• The number tells the magnitude of the energy change (the amount).

• The sign tells the direction of the heat flow.• The direction of heat flow is always from

the system’s point of view.• Positive (+) = heat into the system• Negative (-) = heat out of the system

Exothermic

• Heat flows out of the system into the surroundings.

• q = -x

• system’s energy decreases

• E < 0

Endothermic

• Heat flows into the system from the surroundings.

• q = +x

• system’s energy increases

• E > 0 If you understand the difference between exothermic

and endothermic processes, you just need to remember that heat flow is put in terms of the system

(from the system’s POV) and E will make sense.

Measuring Energy Changes

• 2 common units of energy:

• joule

• calorie

• A calorie is the amount of energy (heat) required to raise the temperature of one gram of water by one Celsius degree.

• The Calories that you count in your food (with a capital C) are actually kilocalories, or 1000 calories.

Measuring Energy Changes

1 Calorie (C) = 1000 calories (cal)

1 calorie (cal) = 4.184 joules (J)

You will need to be able to convert between Calories, calories, and joules.

Practice Problems

1. A small package of fruit snacks has only 80 Calories (kilocalories). How many

calories (cal) are you consuming if you eat all of the fruit snacks in the package?

2. Express 60.1 cal of energy in units of joules.

3. How many calories of energy correspond to 28.4 J?

Practice Problems

Your McDonald’s Meal

Using the information provided, calculate the amount of energy in joules that you consume when you eat your favorite

McDonalds meal. Don’t forget to include the fries and drink!

Calculating Energy Requirements

Section 3


• Remember that 1 calorie is the amount of energy (heat) required to raise the temperature of 1 gram of water by 1oC.

• Remember that temperature only depends on the motion of the particles in a substance, but heat depends on the temperature and the mass of the substance.


• Knowing these facts allows us to calculate the amount of energy required to raise the temperature of a given mass of water a certain number of degrees.

Calculating Energy Requirements Practice Problem 1

Determine the amount of energy (heat) in joules required to raise the temperature of

7.40 g of water from 29.0oC to 46.0oC.


What do we know?

• Mass of water = 7.40 g

• Initial temperature = 29.0oC

• Final temperature = 46.0oC

• 1 cal = 4.184 joules


Where do we want to go?

• 7.40 g water at 29.0oC 7.40 g water at 46.0oC takes how much energy?


How do we get there?

• 1.00 g water at 29.0oC 1.00 g water at 30oC would take 4.184 J of energy.

• Because we have 7.40 g of water instead of 1.00 g, it will take 7.40 x 4.184 J to raise the temperature by 1 degree.

• Thus, 7.40 g at 29oC 7.40 g at 30oC would take 7.40 x 4.184 J of energy.


• However, we want to raise the temperature by more than 1oC. A change from 29.0oC to 46.0oC is a total change of 17.0oC • Final temp. – Initial temp = Total change

in temp• Thus, it will take 17.0 times the energy

necessary to raise the temperature by 1oC.


• In summary, the total amount of energy needed to raise the temperature of 7.40 g of water from 29.0oC to 46.0oC is equal to

17.0 x 7.40 x 4.184

• What are the units?

17.0oC = temperature change

7.40 g = mass of water

4.184 J/g oC = energy per gram of water per degree of temperature


So, how much energy (heat) in joules is required to raise the temperature of 7.40 g

of water from 29.0oC to 46.0oC?

17.0oC x 7.40 g x 4.184 J/g oC = 526 J

Calculating Energy Requirements Review

• The energy (heat) required to change the temperature of a substance depends on:

• The amount of substance being heated (# of grams)

• The temperature change (# of degrees)

• And one other factor we haven’t yet pointed out:

the identity of the substance.

Specific Heat Capacity

• Specific heat capacity is the amount of energy needed to raise the temperature of one gram of a substance by one Celsius degree.

• For water, this amount is 4.184 J because 1 calorie = 4.184 J and it takes exactly 1 calorie of energy to raise the temperature of 1 gram of water by 1oC.

Specific Heat Capacity

• If 4.184 J of energy were added to 1 gram of gold, the temperature would increase by 32oC!

• The specific heat capacity of water is very high, which allows bodies of water to absorb large amounts of heat with very small increases in temperature.


• We can calculate the amount of energy required to raise the temperature of a given amount of a substance using the substance’s mass, the substance’s specific heat capacity, and the temperature change.

• Set it up like the previous practice problem for water!


What quantity of energy (in joules) is required to heat a piece of iron with a

mass of 1.3 g from 25oC to 46oC?


• There is a pattern here! To calculate the energy (heat) required, we multiply the sample size in grams times the specific heat capacity times the change in temperature in Celsius degrees.

Q = msT


Determine the amount of energy as heat that is required to raise the temperature of a 10.0-g sample of aluminum from 25oC to

58oC. Answer in joules and calories.


A 1.6-g sample of a metal that has the appearance of gold requires 5.8 J of

energy to change its temperature from 23oC to 41oC. Is the metal pure gold?

Calorimetry

• A calorimeter is a device used to determine the heat associated with a chemical reaction or physical change.

• Calorimetry experiments determine the enthalpy changes of reactions by making accurate measurements of temperature changes produced in a calorimeter.

Reactions in a Calorimeter

• Q = msT is solved for the water (which is the surroundings)

• The heat (Q) of the reaction (which is the system) will be equal to the heat of the surroundings, but the opposite sign

• If the system gains heat, the surroundings loses the same amount

Energy and Chemical Reactions

Section 4

Enthalpy

• Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.

H = heat given off or absorbed during a reaction at constant pressure

H = Hproducts – H reactants

Thermochemistry

Hproducts < Hreactants

H < 0 H > 0

Hproducts > Hreactants

Enthalpy of Waterhttp://www.mhhe.com/physsci/chemistry/animations/

chang_7e_esp/enm1s3_4.swf

Thermochemical Equations

Is H negative or positive?

System absorbs heat =Endothermic

H > 0

6.01 kJ of energy are absorbed for every 1 mole of ice that melts at 0oC and 1 atm.

H2O (s) H2O (l) H = 6.01 kJ

Thermochemical Equations• The stoichiometric coefficients always refer to the

number of moles of a substance

H2O (s) H2O (l) H = 6.01 kJ

• If you reverse a reaction, the sign of H changes

H2O (l) H2O (s) H = -6.01 kJ

• If you multiply both sides of the equation by a factor n, then H must change by the same factor n.

2H2O (s) 2H2O (l)H = 2 x 6.01 = 12.0 kJ


The physical states of all reactants and products must be specified in thermochemical equations.

H2O (s) H2O (l) H = 6.01 kJ

H2O (l) H2O (g) H = 44.0 kJ


How much heat is evolved when 266 g of white phosphorus (P4) burn in air?

P4 (s) + 5O2 (g) P4O10 (s) H = -3013 kJ

266 g P4

1 mol P4

123.9 g P4

x3013 kJ1 mol P4

x = 6470 kJ

The amount of heat absorbed or released in a chemical reaction depends on the # of moles

reacted.

Practice Problems

When 1 mol of methane is burned at constant pressure, 890 kJ of energy is released as heat. Calculate H for a process in which a 5.8-g sample of

methane is burned at constant pressure.

Practice Problems

How much heat will be released if 1.0 g of hydrogen peroxide (H2O2) decomposes to form water and oxygen in a bombardier beetle to produce a steam spray against

predators?

H = -190 kJ

Practice Problems

The reaction that occurs in the heat packs used to treat sports injuries is

4Fe(s) + 3O2(g) 2Fe2O3(s)

H = -1652 kJ

How much heat is released when 1.00 g of Fe(s) is reacted with excess O2?

And what if…

• Two reactant masses are given in the problem?• Find the limiting reactant by determining

which transfers the smaller amount of energy.

• The reaction is done in a calorimeter?• Calculate qsurroundings (the water)

• Reverse the sign for the qsystem (the reaction)

• Divide the qsystem by the number of moles of reactant used in the reaction

Hess’s Law

Section 5

Hess’s Law

• States that if a series of reactions are added together, the enthalpy change for the net reaction will be the sum of the enthalpy changes for the individual steps.

• In other words, in going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.• Energy is a state function!

Hess’s Law

• The oxidation of nitrogen to produce nitrogen dioxide can occur in one step:

N2(g) + 2O2(g) 2NO2(g) H = 68 kJ

• The same net reaction can also be carried out in two distinct steps:

N2(g) + O2(g) 2NO(g) H = 180 kJ

2NO(g) + O2(g) 2NO2(g) H = -112 kJ

N2(g) + 2O2(g) 2NO2(g) H = 68 kJ

Hess’s Law

• Hess’s Law is useful because it allows us to calculate heats of reaction that might be difficult or inconvenient to measure directly in a calorimeter.

Characteristics of H

• In order to use Hess’s Law, we must review and understand two characteristics of enthalpy changes:1. If a reaction is reversed, the sign of H

is also reversed.2. If the coefficients in a balanced

equation are multiplied by an integer, the value of H is multiplied by the same integer.

Applying Hess’s Law

• Verify that the reactants and products of the net reaction are on the correct sides of the step reactions.

If a step reaction must be reversed, reverse the sign of H for that reaction as well.


• Verify that the coefficients in the net reaction match those in the step reactions.

If a reaction must be multiplied (or divided) by a factor, multiply (or divide) the H value by that factor as well.


• Cancel any elements or compounds that occur as a product in one step reaction and a reactant in another step.

Coefficients must match to cancel completely, otherwise some will be left over

*Note: These general steps are not always done in this order, but should all be applied to Hess’s Law Problems

Practice Problem

The combustion of sulfur can produce SO2 as well as SO3 depending upon the supply of oxygen. From the following reactions and

their enthalpy changes,

2SO2(g) + O2(g) 2SO3 (g) H = -196 kJ

2S(g) + 3O2(g) 2SO3(g) H = -790 kJcalculate the enthalpy change for the combustion of sulfur to produce SO2.

S(g) + O2(g) SO2(g)

Practice Problem

Given the following data:

2O3(g) 3O2 (g) H = -427 kJ

O2(g) 2O(g) H = +495 kJ

NO(g) + O3(g) NO2(g) + O2(g)

H = -199 kJ

calculate the enthalpy change for the reaction

NO(g) + O(g) NO2(g)

Energy Unit 8, Chapter 10. Energy, Temperature, and Heat Section 1.

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