Chemistry 3202Squires Thermo Notes 2013 (Text Ch 16 and 17, page 624)

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Thermochemistry is the study of energy changes that happen when matter changes, when it melts or boils (physical change)

or burns (chemical change). Thermo is greek for hot.

Energy - the ability to do work, en is greek word for in, & ergon is for work. Energon = energy, or work in

This energy can be changed from one form to another, for example, electrical energy can be changed to light.

2 Types of energy:

1) Kinetic = Greek work for movement is kinema (like cinema for moving picture) where temperature is used to describe the average kinetic energy of particles

2) Potential stored energy in the bonds of compounds, or the intermolecular forces between molecules in phase changes .

Heat Flow:

Heat flows from a warmer object to a cooler object, only goes from hot to cold NOT the other way around.

So if we have 2 blocks of Aluminium metal (same mass) one at 80 celcius touching the other one at 20 celcius. Then they will equal out to 50 degress as the hot block losses its energy to the cold one,

NOT the other way around. HOT TO COLD, a one way street.

Energy Units

The unit for energy is the joule (J). Also, 1 kilojoule = 1000 J. Energy can also be measured in calories (still is for food only food is called calories but

Is actually a Kilocalorie. Anyway: 1 cal = 4.184 J )

Physical Change

Example using a Physical Change of state:H2O(l) + energy (40.7 KJ/mol) H2O(g)

The energy, in this case, is absorbed and the reaction is thus endothermic. For the process:

H2O(l) H2O(s) + energy (6.02 KJ/mol) (I call this freezing) The reverse is weirdly called fusion, or melting

In this process, freezing, heat is lost and the phase change is exothermic.

Chemical Change- where the chemicals change

Unlike a physical chage, where the chemicals stay the same, as in melting and boiling.

In a chemical change, one substance(s)(reactants) with a specific amount of energy, is changed into another substance(s)(products) with a different amount of energy.

-Energy changes involve changes in bonding; thus the energy involved is potential energy.

-Changes in potential energy are due to changes in the covalent and ionic bonds.

-For all reactions, energy is absorbed for bonds to break and is released when new bonds form.

Conversion from one form of energy to another obeys the first Law of Thermodynamics, no energy is lost or made when

we change forms of energy, say the chemical change of the burning of coal in a power station into:

Thermal (heat) energy from the burning

Electrical energy from the spinning turbines coils of wire (in a magnetic field) sent to your home.

Mechanical (kinetic + potential) energy of the trbines

Light energy for you to read this by

Chemical energy again , if you are reading this by a battery charged laptop!

Example of a chemical change:

1. Propane burns:

C3H8(g) + 5O2(g) 3CO2 + 4 H2O(g) + Energy (Exothermic Reaction)

2. Photosynthesis (MEMORIZE THIS ENDOTHERMIC REACTION) . It isoften asked.

6CO2(g) + 6 H2O(g) + 3177.6 kJ C6H12O6(l) + 6O2(g)

Photosynthesis is an endothermic reaction due to the reaction requiring 3177.6 kJ of energy for the formation of one mole of glucose. The energy is stored in the glucose as chemical potential energy. Light and chlorophyll are also required. The reaction can also be written OUTSIDE of te equation and then a sign (here positive for Endo thermic) is used:

6CO2(g) + 6 H2O(g) C6H12O6(l) + 6O2(g) H = +3177.6 kJ

The opposite of photosynthesis, cellular respiration (a form of burning), is accompanied by an energy release.

C6H12O6(l) + 6O2(g)6CO2(g) + 6 H2O(g) + 3177.6 kJ

OR

C6H12O6(l) + 6O2(g)6CO2(g) + 6 H2O(g) H = -3177.6 kJ

Thus, in an endothermic reaction, the gaining of heat energy can be written in two different ways:

(1) In the equation, on the reactant side.

(2) Outside the equation as H = +

In an exothermic reaction, the releasing of heat energy can be written:

(1) In the equation, on the product side.

(2) Outside the equation as H = -

Historical Backgrounf to the development of the Joule

Energy is the ability to do work or heat. Physics students calculate work as: W= F x d (m x a x d). This is easy for Physics students to measure, heck , all they need is a meter stick and balance.

In chemistry we have a hard time catching heat, and measuring it is even harder. Heat is not tangible, you cant grab it.

James Joules experiment solves all that. He catches heat and converts the POTENTIAL energy of a raised brick into the KINETIC energy of a moving brick. This mechanically generated value, gives 4.184 J /g C (the mechanical equivalent of heat). (Note: In 1850, Joule published a refined measurement of 772ftlb or Btu (4.1 J/cal) or 772 pounds raised 1 foot generates 1 Fahrenheit temperature change in 1 pound of water. We use Metric today or 4.184 J/g C

The First Calorimeter

Temperature Meaasures Kinetic Energy

Temperature (T): a measure of the average kinetic energy of the particles. Or a thermometer is a molecular speedometer.

Molar Heat of Combustion

Burning one mole of a substance is called the molar heat of combustion. The candle lab is a good example.

Another Example: The molar heat of combustion of methane gas is 890.4 kJ.

CH4(g) + 2O2(g) CO2(g) + H2O(g) H = -890.4 kJ

Systems, Surrounding, and Universe

The First Law of Thermodynamics

Energy is neither created nor destroyed, it is transformed. Also known as the Law of Conservation of Energy

Or, more simply, heat is transferred from a hotter object to () a colder object, again, a one way street.

The Second Law of Thermodynamics

A rock will fall if you lift it up and then let go. Hot frying pans cool down when taken off the stove. Iron rusts (oxidizes) in the air. Air in a high-pressure tire shoots out from even a small hole in its side to the lower pressure atmosphere. Ice cubes melt in a warm room. No matter how often you organize your stuff, your room, it eventually just ends up cluttered again!

Whats happening in each of those processes? Energy of some kind is changing from being localized ("concentrated" in the rock or the pan, etc.) to becoming more spread out. Look at those examples again to see how that statement fits them all.

A simple way of stating fundamental science behind the second law: Energy spontaneously disperses from being concentrated (or localized) to becoming spread out if it is not hindered from doing so.

Entropy means being being scattered or disordered. A rock has potential energy localized in it when you lift it up above the ground. Drop it and that PE changes to kinetic energy (energy of movement), pushing air aside as it falls (therefore spreading out the rocks KE a bit) before it hits the ground, dispersing a tiny bit of sound energy (compressed air) and causing a little heating of the ground it hits and the rock itself. The rock is unchanged (after a minute when it disperses to the air the small amount of heat it got from hitting the ground).But the potential energy that your muscles localized in by lifting it up is now totally spread out and dispersed all over in a little air movement and a little heating of the air and ground.

Systems and Surroundings

System: The part being studied, the chemical.

Surroundings: The parts of the universe, the surroundings, with which the system is in contact.

Types of Systems

1. Open Systems:

Both energy and matter can be transferred between a system and its

surroundings. Example: An open beaker of hot water loses heat to the surroundings

upon cooling loses matter in the form of water vapor.

System = water molecules

Surroundings = beaker + air near it

2. Closed Systems:

Energy can be exchanged between the system and its surroundings, but matter cannot.

Example: A stoppered erlenmeyer flask of hot water loses heat to the surroundings, however, water vapor cannot escape.

System = water molecules

Surroundings = beaker + stopper + air near it

3. Isolated Systems:

Neither energy nor matter is exchanged between the system and its surroundings.

Thermally isolated, well now come to think of it, TOTALLY isolated.

Example: A vacuum flask or vacuum thermos of hot water approximates an isolated system.

SURROUNDINGS

System = liquid contents

Surroundings = thermos + air near it

A vacuum contains no matter, it cannot transfer heat by conduction or convection.

PHASE CHANGES

A change in the state of matter without any change in the chemical compostion. Usually that is melting or boiling, however PHYSICAL changes may also be things such as sublimation, which occurs when you go from solid to liquid, quickly skipping through the liquid phase as with :

I2 (s) I2 (g) or dry ice, in smoke machines.. CO 2(s) dry ice CO 2 (g)

PHYSICAL changes may also be simply dissolving NaCl (s) NaCl (aq) which further dissociates to Na + (aq) + Cl - (aq)

In the lab we disslove NaOH all the time and you remember it gave of a LOT of heat.

NaOH(s) Na + (aq) + OH - (aq) + ENERGY

However, the major calculations, for physical changes, we do will mainly involve Phase changes ( and the Kinetic changes that bring us to the melting or boiling point).

Thus there is no change in the average kinetic energy of the particles for a phase change, they never involve temperature changes. Temperature changes are for KINETIC (mcT and CT) ONLY.

A change in potential energy is taking place due to changes in theattractions between particles (IMF, intermoloceular forces, for molecular substances).

Heating Curve for Water

GENERAL STEPS:

Leg A.

Heating of Solid: As the temperature increases, kinetic energy increases and potential energy remains constant.

The energy goes to Increased vibration of particles about their fixed positions in solid.

Leg B.

Fusion (Melting) of Solid: Here the temperature remains constant and there is no change in the kinetic energy,

however the potential energy increases. The added energy weakens the attractions between the particles, so

they move more freely; now the substance changes into a liquid

Leg C.

Heating of Liquid: The temperature again rises causing an increase in kinetic energy

but the potential energy remains constant.

The particles moving more freely, but the attractions are still strong enough to hold particles in the liquid phase.

Leg D.

Vaporization (Boiling) of Liquid: Again, like #B, this is a phase change so

the temperatue remains constant. This means the kinetic energy is constant

and the potential energy is increasing.

Again, the energy is used to overcome the attractions between particles as the particles break free

and the substance changes into the gas phase.

Leg E.

Heating of Gas: The temperature begins to rise above its boiling point causing the kinetic energy to increase

and potential energy to remain constant. Adding heat speeds up the gas particles.

Calculating Kinetic Energy:

Kinetic energy involves movement of molecules, thus there will be a temp change as a thermometer is a molecular speedometer.

q = mcT where q = the amount of energy in Joules or KJ

m = the mass of the system (usually only the mass of water)

c = the specific heat of the system (a constant for each material)

T = the change in temperature

Molar Heat

The total energy of a system is called the enthalpy (H) of the system, also referred to as Molar Enthalpy ( H):

q = n H ( the enthalpy change involving one mole of a substance, the units = kJ/mol. )

Sometimes, the heat of the reaction may be expressed in terms of a molar enthalpy (kJ/mole).

Hrx = q / # of moles or Hrx = mcT / # of moles

Example: Calculate the heat required to melt 50.0 g of ice at 0.0C.

Answer

Notice the sign of the heat value. Melting (fusion) ice

is an endothermic process - energy is required to break the intermolecular forces that hold water molecules together in the ice crystal. The energy absorbed by the ice becomes the potential energy of the liquid molecules.

Types of Problems

1) Physical Changes

Ex: A red hot horseshoe is dropped into 1000.0 g ( always be sure to check the units of mass as they must match the units of specific heat, KJ with KJ, and J with J ) of water. If the temperature of the water rose from 22.1 oC to 35.9 oC, how much heat energy did the water gain?

q = mcT

m = 1000.0 g

c = 4.18 J g-1 oC-1

T = (35.9 oC 22.1 oC) = 13.8 oC

q = (1000.0 g)( 4.18 J g-1 oC-1)( 13.8 oC)

q = 57.7 kJ or 5.77 x 104 J

q = 57684 J

Another Example:

You put a 500.0 mL bottle of water in the fridge for 2.0 hours and the temperature of the water drops from 22.0 oC to 5.0 oC. How much heat energy was removed from the water?

q = mcT

m = 500.0 mL x 1.0 g mL-1 = 500.0 g (or 500ml is 500g for water)

c = 4.18 J g-1 oC-1

T = l (5.0 oC 22.0 oC) l

= 17.0 oC

q = (500.0 g)( 4.18 J g-1 oC-1)( 17.0 oC)

q = 35.5 kJ or 3.55 x 104 J

NOTE - this is an exothermic change as heat lost to fridge. Practice Problems: Page 636: #s 7 10

2) Chemical Changes

Example: A student placed 50.0 mL of 1.00 M HCl at 25.5 oC in a coffee cup calorimeter. To this, 50.0 mL of 1.00 M NaOH also at 25.5 oC was added. The mixture temperature rose to a maximum of 32.4 oC. Calculate the heat of the reaction (heat of neutralization).

mAcid = 50.0 mL x 1.0 g mL-1 = 50.0 g we also have 50 grams base OR 100 grams acid and base which when reacted is water.

c = 4.18 J g-1 oC-1

T = (32.4 oC 25.5 oC) = 6.9 oC

q = mcT

q = (100.0 g water (acid and base combo))( 4.18 J g-1 oC-1)( 6.9 oC)

q = 2.88 kJ

H per mole of HCl

n = C x v

n = ( 1.00 M )( 0.0500 L)

n = 0.0500 mole HCl

H = - 2.88 kJ/ 0.0500 mole HCl

= - 57.7 kJ mole-1

Wasnt that funInstead of using the specific heat of water, we look at the specific heat of the whole calorimeter heat capacity.

Heat Capacity (C):

The quantity of heat required to change the temperature of an object by one degree Celsius, or q = Ct (really the same as q = mct , just that mc are combined

or already multipied for you)

specific heat (c) = q m-1 T-1 the amount of heat needed to

= J g-1 oC-1 increase 1 g of material by 1 oC

whereas:

heat capacity (C) = q T-1 the amount of heat needed to

= J oC-1 increase an object by 1 oC (notice mass is absent or incorporated into C)

Example:

1.32 g of sucrose, C12H22O11, is burned in a bomb calorimeter with a heat capacity of 9.43 x 103 J oC-1.

If the temperature of the calorimeter changed from 25.00 oC to 27.31 oC, calculate the heat of the reaction in units of kJ mole-1 of sucrose.

C = 9.43 x 103 J oC-1

T = 2.31 oC

q = C x T

q= 21.8 kJ

moles sucrose = mass / molar mass

= 1.32 g C12H22O11 / 342 g . mole-1

= 3.8596 x 10-3 mole C12H22O11

H = - 21.7833 kJ / 3.8596 x 10-3 mole C12H22O11

H = - 5643.8 kJ mole-1

H = - 5.64 x 103 kJ mole-1

There are 3 types of changes in Thermodynamics you need to know:

We are concerned with 3 types of changes in:

a) Physical changes - Tens of KJ/mol

b) Chemical Changes - Hundreds of KJ/mol

c) Nuclear changes - Millions and even Billions KJ/mol

physical

H2(l) H2(g)

100 to 102 kJ/mole

(0 to 100)

chemical

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

102 to 104 kJ/mole

(100 to 10,000)

nuclear

Hnucleus + Hnucleus Henucleus

1010 to 1012 kJ/mole

Millions to Billions

CHEMICAL CHANGE CALCULATIONS

Knowing the heat change associated with a chemical change involving one mole of a chemical allows you to calculate the heat change associated with any measured amount of that chemical.

Example : Calculate the heat change that should occur when 10.4 g of CaCl2 is formed from its elements . The molar heat for this FORMATION reaction can be obtained from molar heats of formation tables, but is given here as -795.4 KJ/mol. Note: THIS is a CHEMICAL change, or a reaction.

Ca(s) + Cl2 (g) CaCl2 (s)

Example : A 5.00 g sample of benzene was burned at SATP conditions resulting in the release of 209.2 kJ of heat. Calculate the molar heat of combustion for benzene.

The molar heat of combustion of benzene is -3267 kJ/mol.

Heat of Reaction and Simple Calorimetry

Heat changes associated with aqueous chemical reactions can be determined by simple calorimetry. This information can then be used to calculate heat of reaction.

HCl + NaOH HOH + NaCl

Example:

A 75.00 mL sample of 0.500 M hydrochloric acid at 25.00C was mixed with 75.00 mL of 0.500 M NaOH in a simple calorimeter. The highest temperature recorded after mixing was 27.19C. Calculate the molar heat of reaction for hydrochloric acid.

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CHEMICAL CHANGE CALCULATIONS WITHOUT CALORIMETERS

1) Hess's Law: (FLIP THE EQUATION METHOD)

The basis for determining an enthalpy change by calculation is an idea put forth by Mr. Hess. It states, if two or more thermochemical equations can be added together to produce an overall equation, the sum of the enthalpy terms of the thermochemical equations equals the enthalpy change of the overall equation. The manipulation of thermochemical equations is the key to applying the law of heat summation.

Ex: Given these thermochemical equations:

use Hesss law of heat summation calculate the molar enthalpy for the incomplete combustion of graphite:

2) Hess's Law: (THE MATH METHOD)

There is a second way to use Hess' Law :

Hrxn = nHf, products minus n Hf, reactants

Problem #1: Calculate the standard enthalpy of combustion for the following reaction:

C2H5OH (l) + (7/2) O2 (g) ---> 2 CO2 (g) + 3 H2O (l)

The Hf values I will now give as, FOR 1 Mole:

-393.5 for CO2,

-286 for water,

-278 for ethanol

zero for ALL elements

The key to solving this problem is to have a table of standard enthalpies of formation handy.

Like this:

Hcombustion = [ 2 (-393.5) + 3 (-286) ] minus [ (-278) + (7/2) (0) ]

Doing the math gives us Hcomb = -1367 kJ/mol of ethyl alcohol.

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3) Bond Energies.

The process of measuring the changes in the enthalpy of a system is called calorimetry. The change in enthalpy, H, is related to what is happening in the reaction.

Ex: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)

We can burn methane and the heat it gives off can be absorbed and measured by a cup of water (calorimeter)

Bonds being broken (THIS ABSORBS ENERGY): : CH4(g) + 2 O2(g) Products

4 C-H bonds are being broken both of

2 O-O bonds are being broken these require energy

Bonds being formed (THIS RELEASES ENERGY): Reactants CO2(g) + 2 H2O(l)

2 C=O bonds being formed both of

4 H-O bonds being formed these release energy

As the type and number of bonds being broken and formed change, so will the change in enthalpy.

H = Hbond energy reactants - H bond energy products

Calculating Bond Energies:

Bond breaking is endothermic, because energy is absorbed to separate bonded particles. The potential energy of the separated particles is greater than that of the bonded particles.

The amount of energy needed to separate one mole of a bonded species is called the bond energy. Keeping the sign convention in mind, all bond energies are positive!

Example: Using average bond energies, calculate the enthalpy change for the combustion of ethanol.

Enthalpy Diagrams

Potential energy diagrams, also known as enthalpy diagrams, show the relative potential energies of the species involved in a chemical (or physical) change. Molar enthalpy is the potential energy change associated with a physical or chemical change involving one mole of a substance. You may heard this as the heat of reaction. Consider the following EXOTHERMIC diagram.

The arrow pointing downwards indicates the exothermic nature of the reaction. The potential energy of the reactants is greater than that of the products. The enthalpy of the system decreases as potential energy is released to the surroundings as heat.

Now consider this ENDOTHERMIC diagram.

The enthalpy diagram for an endothermic reaction always shows the arrow pointing upwards. The potential energy of the system increases as energy is absorbed from the surroundings.

ALL TOPICS

SAMPLE PROBLEMS:

Problem #1: How many kJ are required to heat 45.0 g of H2O at 25.0 C and then boil it all away?

Solution:

Comment: We must do two calculations and then sum the answers.

1) The first calculation uses this equation:

q = (mass) (t) (Cp)

This summarizes the information needed:

t = 75.0 CThe mass = 45.0 gCp = 4.184 J g1 C1

2) Substituting, we have:

q = (45.0 g) (75.0 C) (4.184 J g1 C1) so q = 14121 J = 14.121 kJ

3) The second calculation uses this equation:

q = (moles of water) (Hvap)

This summarizes the information needed:

Hvap = 40.7 kJ/molThe mass = 45.0 gThe molar mass of H2O = 18.0 gram/mol

4) Substituting, we obtain:

q = (45.0 g / 18.0 g mol1) (40.7 kJ/mol)

q = 101.75 kJ

5) Adding:

101.75 kJ + 14.121 kJ = 116 kJ (to three sig figs)

Problem #2: How many kJ need to be removed from a 120.0 g sample of water, initially at 25.0 C, in order to freeze it at 0 C? (Area three, then area two on the time-temperature graph.)

Solution:

1) The first calculation:

q = (mass) (t) (Cp)

q = (120.0 g) (25.0 C) (4.184 J g1 C1)

q = 12,552 J = 12.552 kJ

2) The second calculation:

q = (moles of water) (Hvap)

q = (120.0 g / 18.0 g mol1) (6.02 kJ/mol)

q = 40.13 kJ

3) Summing up the values from the two steps gives 52.8 kJ.

Problem #3: You are given 12.0 g of ice at -5.00 C. How much energy is needed to melt the ice completely to water?

Solution:

1) The first calculation:

q = (mass) (t) (Cp)

q = (12.0 g) (5.0 C) (2.06 J g1 C1)

q = 123.6 J = 0.1236 kJ

2) The second calculation:

q = (moles of water) (Hvap)

q = (12.0 g / 18.0 g mol1) (6.02 kJ/mol)

q = 4.0133 kJ

3) Summing up the values from the two steps gives 4.14 kJ, to three significant figures.

Problem #4: Equal masses of hot water and ice are mixed together. All of the ice melts and the final temperature of the mixture is 0 C. If the ice was originally at 0 C, what was the initial temperature of the hot water?

Solution:

Let us assume we have 18.0 g of ice and 18.0 g of hot water present.

The key is to realize that the only thing the ice did is melt, it did not change its temperature. So, let us calculate the amount of heat needed to melt our 18.0 g (or, 1.00 mole) of ice:

q = (6.02 kJ/mol) (1.00 mol) = 6.02 kJ

The only source of heat is the hot water, which provides 6020 J (I converted the 6.02 kJ to J.) of heat. Let us calculate the temperature change of 18.0 g of hot water as it loses 6020 J of heat:

6020 J = (18.0 g) (x) (4.184 J g1 C1)

x = 79.9 C

The hot water was at an initial temperature of 79.9 C (since everything ended up at a final temperature of 0 C.

Three Equations Needed

Problem #5: A 36.0 g sample of water is initially at 10.0 C. How much energy is required to turn it into steam at 200.0 C? (This example starts with a temperature change, then a phase change followed by another temperature change.)

Solution:

q = (36.0 g) (90.0 C) (4.184 J g1 C1) = 13,556 J = 13.556 kJ

q = (40.7 kJ/mol) (36.0 g / 18.0 g/mol) = 81.4 kJ

q = (36.0 g) (100.0 C) (2.02 J g1 C1) = 7272 J = 7.272 kJ

q = 102 kJ (rounded to the appropriate number of significant figures)

Problem #6: A 72.0 g sample of ice is at 0 C. How much energy is required to convert it to steam at 100.0 C? (This example begins with phase change, then a temperature change and then a second phase change, areas two, three and four on the time-temperature graph.)

Solution:

q = (6.02 kJ/mol) (72.0 g / 18.0 g/mol) = 24.08 kJ

q = (72.0 g) (100.0 C) (4.184 J g1 C1) = 30,125 J = 30.125 kJ

q = (40.7 kJ/mol) (72.0 g / 18.0 g/mol) = 162.8 kJ

q = 217 kJ (rounded to the appropriate number of significant figures)

Problem #7: Calculate the heat released by cooling 54.0 g H2O from 57.0 C to minus 3.0 C. (Cools from 57.0 to zero, a phase change, the cools from zero to -3; three equations needed.)

Solution:

q = (54.0 g) (57.0 C) (4.184 J g1 C1) = 12,878 J = 12.878 kJ

q = (6.02 kJ/mol) (54.0 g / 18.0 g/mol) = 18.06 kJ

q = (54.0 g) (3.0 C) (2.06 J g1 C1) = 333.73 J = 0.334 kJ

q = 31.3 kJ (rounded to the appropriate number of significant figures)

Problem #8: How much energy is required to heat 125.0 g of water from -15.0 C to 35.0 C?

The solution:

q = (125.0 g) (15.0 C) (2.06 J g1 C1) = 3862.5 J = 3.8625 kJ

q = (6.02 kJ/mol) (125.0 g / 18.0 g/mol) = 41.806 kJ

q = (125.0 g) (35.0 C) (4.184 J g1 C1) = 18,305 J = 18.305 kJ

q = 64.0 kJ (rounded to the appropriate number of significant figures)

Problem #9: 18.0 mL of water at 28.0 C are added to a hot skillet. All of the water is converted to steam at 100.0 C. The mass of the pan is 1.25 kg and the molar heat capacity of iron is 25.19 J/mol C. What is the temperature change of the skillet?

Solution:

1) Determine the energy need to heat and boil the water:

heat: q = (18.0 g) (72.0 C) (4.184 J/g C) = 5422.464 J = 5.422464 kJ

boil: q = (40.7 kJ/mol) (18.0 g / 18.0 g/mol) = 40.7 kJ

total: 40.7 kJ + 5.422464 kJ = 46.122464 kJ

I won't bother to round off until the final answer.

2) Determine temperature change of skillet:

46122.464 J = (22.38338 mol) (t) (25.19 J/mol C)

t = 81.8 C

Note that moles of iron are used rather than grams. This is because of the units on the specific heat provided in the problem.

Problem #10: A 10.35 kg block of ice has a temperature of -22.3 C. The block absorbs 4.696 x 106 J of heat. What is the final temperature of the liquid water?

Solution:

1) Raise the temperature of the ice to 0 C:

q = (10350 g) (22.3 C) (2.06 J/g1 C1)

q = 475458.3 J

2) Melt the ice at 0 C:

q = (6020 J/mol) (10350 g / 18.015 g/mol)

q = 3458617.818 J

3) How many Joules remain?

total used to this point = 3.934 x 106 J

total remaining: 4.696 x 106 J minus 3.934 x 106 J = 7.62 x 105 J

4) Determine temperature increase of liquid water:

7.62 x 105 J = (10350 g) (x) (4.184 J/g1 C1)

x = 1.76 C

Problem #11: How many grams of ice at -5.6 C can be completely converted to liquid at 10.1 C, if the available heat for this process is 4.74 x 103 kJ?

Solution:

1) Raise the temperature of the ice to zero Celsius:

q = (x) (5.6 C) (2.06 J/g1 C1)

2) Melt the ice at zero Celsius:

q = (6020 J/mol) (x / 18.015 g/mol)

3) Raise the temperature of the ice to 10.1 Celsius:

q = (x) (10.1 C) (4.184 J/g1 C1)

4) 4.74 x 106 J are used in total:

4.74 x 106 J = [(x) (5.6 C) (2.06 J/g1 C1)] + [(6020 J/mol) (x / 18.015 g/mol)] + [(x) (10.1 C) (4.184 J/g1 C1)]

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