Chemistry is about to heat

up…Please get out:1. Last Large

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THERMOCHEMISTRY

Chemical Potential Energy Energy in chemical bonds

Thermal Energy Kinetic energy of molecules This has to do with heat!

HEAT What this part is about

TYPES OF ENERGY

Heat will always flow from a hot substance to a cold substance

Temperature is not Heat

Temperature Average kinetic energy

Heat A form of energy

Think of a car driving down the road A car has a temperature, but it also has kinetic energy We do not say ‘The car is going 77°F’ Just like we don’t say ‘The jar has 40°C of heat’

HEAT

Endothermic Takes energy (usually cold) Will absorb energy from surroundings

Exothermic Releases energy (usually hot) Will release energy to surrounding

EXOTHERMIC/ENDOTHERMIC

System What we are concerned with

Surroundings Everything else

Is this exothermic or endothermic?

Endothermic! The water in the pot absorbs heat from the burner

underneath it

SYSTEM AND SURROUNDING

System

Many Xbox 360’s overheated and broke. Identify the system and surrounding Identify Exothermic or Endothermic

The Xbox releases energy to thesurroundings

SYSTEM VS SURROUNDINGS

System

You are probably familiar with one of the most popular: Calories Note the big ‘C’

A Calorie is a food calorieA calorie is the amount of energy required to heat 1 gram of water 1°C

A Calorie is 1,000 calories Note the little ‘c’

Saying “Two tablespoons of Nutella has 200 Calories is really saying Nutella has 200,000

calories (!)

UNITS

The unit we will use most often is a Joule (J) Or kiloJoule, 1 kJ = 1000 J

4.184 Joules = 1 calorie Conversion Time!How many Joules are in the 200,000 calories of

Nutella?

Or… 836,800 Joules = 836.8 kJ That’s the same amount of energy as me moving at 319

mph

MORE UNITS

Heat Capacity (J/°C) The heat required to raise the temperature 1°C

Specific Heat (J/g*°C) Heat required to raise the temperature of 1 gram of

something 1°C EVERY substance has a different Specific Heat

SPECIFIC HEAT

Specific Heat of Wood

Specific heat of Steel

2 J/g*°C 0.49 J/g*°C

Remember this equation

q=mCΔT Heat equals mass*Specific Heat*Change in Temperature

Or:

Specific Heat equals heat divided by mass and Change in Temperature

CALCULATING HEAT

JoulesOrCalories

grams

𝐽𝑜𝑢𝑙𝑒𝑠𝑔𝑟𝑎𝑚∗℃

𝑜𝑟𝐶𝑎𝑙𝑜𝑟𝑖𝑒𝑠𝑔𝑟𝑎𝑚∗℃ ℃

84 grams of Labradorite absorbs 1008 Joules of energy and the temperature increases by 15°C. What is the Specific Heat of Labradorite?

C=C = 0.8 J/g*°C

CALCULATING SPECIFIC HEAT

Mass Heat

Change in Temperature

2 kg of aluminum is heated from 20oC to 100oC. The specific heat of aluminum is 0.91 kJ/kg0C.

How much heat is required?q=mCΔTq=(0.91 kJ/kg0C)(2 kg)(1000C-200C)

q= 145.6 kJ

MORE CALCULATIONS

Mass

Change in Temperature

Specific Heat

200 grams of copper is heated from 25 to 700. How much heat would that take? The specific heat of copper is 0.386

q=mCΔTq=(200 g) (0.386 ) (7000C-250C)

q = 52,110 JoulesOr q = 52.11 KiloJoules

CALCULATING HEAT

Mass

Change in Temperature

Specific Heat

If it takes about 68 seconds to hair dry 113 grams (average women’s hair mass) of hair, and you use 102 kJ, what is the temperature change of the hair? The specific heat of hair is 0.795 (That was hard to find)

q=mCΔTSolve for ΔT

= 1,135.4˚CWhy doesn’t this happen?We are getting rid of water too – Changing state!

MORE FUN

Mass

Specific Heat

Heat

Does it take energy to change state? Of course!

Heat of fusion/solidification Solidification = liquid to solid Fusion = solid to liquid

Does it take more energy to melt an ice cube than to freeze and ice cube? It’s the same!

Heat of fusion (ΔH fus) = -Heat of solidification (ΔHso l id)Freezing an ice cube Melting an ice

cube

CHANGE OF STATE

333.56 J/g

-333.56 J/g

How much energy is required to melt 225 grams of ice? Hf = 333.56 J/g

Instead of q=mCΔT We just need q = mH f

q = 75,051 Joules

75.051 kJ required

Easy!

I DO

Melting KClO3 requires heat! How much heat is required to melt 22 grams of potassium chlorate? The heat of fusion is 278 J/g

q= mH f

q=6,116 Joules

WE DO

*

Vaporization and condensation work the same way. Vaporization = liquid to vapor Condensation = vapor to liquid

Heat of vaporization = -Heat of condensation ΔHvap = -ΔHcond

What is this energy from? Changing state means increasing the

kinetic energy of the molecules enough to overcome the intermolecular bonds.

Or decreasing the kinetic energy enough such that the intermolecular bonds can hold the molecules together

JUST LIKE BEFORE

I burned some money. Why was the money still there?

There was ethanol on it! Sneaky

The heat of vaporization of Ethanol is 838.26 J/gram. How much heat did the 0.84 grams of ethanol give off ?

q= mHv

q = (0.84 grams)(838.26 )q = 704.14 Joules

BURNING MONEY

It’s about to heat upREMEMBER

Temperature Change

Phase Change

COMBINING PHASE CHANGE + TEMP CHANGE

You’re drink has a 40 gram ice cube (0°C) in it. After 20 minutes the temperature of the ice is 25°C (oh no!). How much heat did the ice gain? The heat of fusion of ice is 333.56 J/gram The specific heat of water is 4.184 J/g*°C

Two Steps Which steps?

Look at this:Key words:

Ice (0°C) 25°C

MELT AND HEAT

Start: Ice at 0˚C

End: Water at 25 ˚C

1) Melt REMEMBER: q = (40 grams)*( 333.56 j/g)

q1= 13342.4 joules

2) Heat q=mCΔT q = (40 grams)(4.184 J/g*°C)(25°C - 0°C) q2 = 4184 joules

TOTAL = Add them up: q1 + q2 = q13342.4 Joules + 4184 Joules = 17526.4 Joules (or 17.5 kJ)

DA MATH

1) Melt

2) Heat

2 Liters (2 kg) of water at 25˚C is heated to 100˚C, vaporized, and then heated to 120˚C. How much energy is needed? Cwater=4.18 Hv = 2257 Csteam = 1.89

Think about what steps are needed.Start: Liquid water at 25˚C →Vaporize → Vapor at

120˚CDraw a small heating curve:Every line is an equation

STEAM

25˚C

100˚C

100˚C

120˚C

1 ) Heat

2 ) Boil 3 ) Heat

1) Heat

q=mCΔT q1 = (2000 grams) q1 = 627,600 Joules (or 627.6 kJ)

2) Boil q = mHv

q2 = (2000 grams) q2 = 4,514,000 Joules (or 4,514 kJ)

3) Heat q=mCΔT q3 = (2000 grams) q3 = 75,600 Joules (or 75.6 kJ)

Add them up!

THE MATH

∗(4.184𝐽𝑜𝑢𝑙𝑒𝑠

𝑔𝑟𝑎𝑚∗℃)∗(100℃−25℃ )

∗(2257𝐽𝑜𝑢𝑙𝑒𝑠𝑔𝑟𝑎𝑚

)

∗(1.89𝐽𝑜𝑢𝑙𝑒𝑠

𝑔𝑟𝑎𝑚∗℃)∗(120℃−100℃)

627.6𝑘𝐽 +4,514𝑘𝐽 +75.6 𝑘𝐽=5,217.2𝑘𝐽