energy and temperature

Energy and Temperature

I. The Nature of Energy

• The ability to do work

• 3 categories

-Radiant energy

-Kinetic energy

-Potential energy

Kinetic Energy

• Energy of motion

• Types

1. Mechanical - moving

parts of a machine

2. Thermal (heat) - energy

caused by the random

internal motion of

particles of matter

Potential Energy

• Stored energy

• Types

1. Electrical - ex. battery

2. Gravitational - ex. water

behind dam - used for

electricity

3. Chemical - ex. chemical

bonds in food

Law of Conservation of Energy

• In any process, energy is neither created nor destroyed.

ex. 1 hitting a baseball transfers kinetic energy from the bat to the ballex. 2 igniting a match changes chemical energy into heat and light

TEMPERATURE VS. HEAT

Temperature• A measure of the average

kinetic energy of molecules in motion

• Remember: Kelvin = 273 + Celsius

Heat• Total amount of energy that flows

between matter• Flows from matter of higher

temperature to matter of lower temperature

• “Hot” molecules quickly move into areas of slower moving “cold” molecules

Visual Concepts

Temperature and the Temperature Scale

Chapter 10

Heat

• The transfer of kinetic energy from a hotter object to a colder object.

• Symbol is q

-particles are always moving

-when you heat water molecules

move faster

EXOTHERMIC REACTIONS

• Chemical reactions that release thermal energy• Feels hot – temperature rises• Examples: condensation, freezing

ENDOTHERMIC REACTIONS

• Chemical reactions that absorb thermal energy• Feels cold – temperature drops• Examples: boiling, evaporation, melting

Measuring Heat• Kelvin Scale (K)

– 0 K - point at which there is no molecular motion (absolute zero)

– All Kelvin temperatures are positive

• K = °C + 273

What is the boiling point of water in Kelvin?

What is the freezing point of water in Kelvin?

Measuring Heat

• Calorie = amount of heat needed to raise the temperature of 1 gram of water by 1 degree Celsius

(1 cal = 1 g x 1 C°)

• Energy stored in food = Calorie (Cal)

1 Cal = 1000 cal = 1 kcal

Joule (J) : 1 cal = 4.184 J

Specific Heat

• Specific Heat– Amount of heat

required to raise 1 gram of a substance 1°C.

– physical property

• Liquid water 4.184 J/g°C

• Fe 0.449 J/g°C

Specific Heat • Water has high heat capacity:

– Absorbs a large quantity of heat with only a small increase in temperature

– Gives up a large quantity of heat with only a small decrease in temperature

• Metals have low heat capacity–Small amount of heat large

temperature change

HEAT

q = cp• m • ΔT

Where

q = heat released (-) or heat absorbed (+)

cp = specific heat (value given on p. 343)

m = mass

ΔT (means change in Temperature)

= Final Temp – Initial Temp

HEAT

q = cp• m • ΔT

Where

q = Joules (J) or calories (cal)

cp = J/g•K or J/g•˚C or cal/g •˚C

m = grams

ΔT = K or °C

SAMPLE PROBLEM A

If 75 g of iron (cp = 0.449 J/g•K) is heated from 274 K to 314 K, what is the amount of heat that is absorbed?

SAMPLE PROBLEM A

If 75 g of iron (cp = 0.449 J/g•K) is heated from 274 K to 314 K, what is the amount of heat that is absorbed?

Step 1:Outline what you know.q = ?m = 75 gcp = 0.449 J/g•KΔT = 314 - 274

SAMPLE PROBLEM A

If 75 g of iron (cp = 0.449 J/g•K) is heated from 274 K to 314 K, what is the amount of heat that is absorbed?

Step 1:Outline what you know.q = ?m = 75 gcp = 0.449 J/g•KΔT = 40 K

SAMPLE PROBLEM A

If 75 g of iron (cp = 0.449 J/g•K) is heated from 274 K to 314 K, what is the amount of heat that is absorbed?

Step 2:Plug-in and solve.

q = m • cp • ΔT

q = (75) (0.449) (40)

q = 1347 J

Endothermic

SAMPLE PROBLEM B

A 4.0 g sample of glass was heated from 274 K to 314 K and was found to have absorbed 32 J of energy as heat. What is the specific heat of this type of glass?

SAMPLE PROBLEM B

A 4.0 g sample of glass was heated from 274 K to 314 K and was found to have absorbed 32 J of energy as heat. What is the specific heat of this type of glass?

Step 1:Outline what you know.q = 32 Jm = 4.0 gCp = ?ΔT = 314 – 274 = 40 K

SAMPLE PROBLEM B

A 4.0 g sample of glass was heated from 274 K to 314 K and was found to have absorbed 32 J of energy as heat. What is the specific heat of this type of glass?

Step 2:Plug-in and solve.

q = m • cp • ΔT

32 = (4.0) (x) (40)

32 = 160 x

x = 0.20 J / g • K

Practice Problems How much heat would be required to

raise the temperature of 20.0 g of water from 50.0ºC to 100. ºC. (c of H2O = 4.18J/gºC)

q = cm ΔT

= (4.18J/gºC)(20.0g)(100. ºC–50.0ºC)

q = 4,180 J

II.A. Calorimetry

• Calorimeter - an insulated device to measure temperature changes

• Can determine enthalpy changes (heat) of reactions

Calorimetry

• Exothermic process releases heat temperature of surroundings increases

• Endothermic process absorbs heat temperature of surroundings decreases

Calorimeter

• qrxn = - qsur

Transferred Surroundings

Put hot iron ring into

cool water

Leave until the temp

Is the same for both

How does heat lost by

the iron compare to

heat gained by the

water?

Heat lost

Heat gained

Water

Iron

LAB: Heat and Molecular MotionPurpose: To illustrate what happens when a

substance is heated.Procedure:1. Fill 2 250 mL beakers ¾ full of tap water.2. Heat 1 to near boiling on a hot plate.3. Gently put 1 drop of food coloring in

each.

Quiz 12-9 A 10.0 g sample of metal X requires

25.0 J of heat to raise its temperature from 17ºC to 67ºC. What is the specific heat of the metal?

Quiz 12-8

How much heat is required to raise the temperature of 30.0 g of H2O from 18.0°C to 28.0°C?

c = 4.18 J/gºC

THERMOCHEMISTRY-

• Study of the changes in heat in chemical reactions

• All chemical reactions involve changes in energy.

• Heat energy is either absorbed or released.

Enthalpy

• Enthalpy (H)-heat content of a substance

• Depends on temperature, physical state, and composition

• Enthalpy Change - the amount of heat absorbed or released during a chemical reaction; ΔH

ΔH = Hproducts – Hreactants

Exothermic Reactions

• Reactions that release heat to their surroundings

• Combustion

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) + heat

- Heat is produced because the energy released as new bonds are formed (products) > energy required to break the old bonds (reactants)

Thermite Reaction

Al(s) +Fe2O3(s) Al2O3(s) +Fe(l) +heat

Exothermic Reaction

• Products have

lower potential

energy than reactants.

Al(s) +Fe2O3(s) Al2O3(s) +Fe(l) +heat

energyreleased

• If heat is released,

• Hproducts < Hreactants

H is negative (exothermic)

Endothermic Reactions

• Reactions that absorb heat from surroundings

• Ammonium nitrate in water

NH4NO3(s) + heat NH4+(aq) + NO3

-(aq)

• Energy released as new bonds are formed (products) < energy required to break bonds (reactants)

• This energy must be supplied by surroundings and is stored in the chemical bonds of the products

Endothermic Reaction

• Reactants have

lower potential energy than products

NH4NO3(s) + heat NH4+(aq) + NO3

-(aq)

energyabsorbed

• If heat is absorbed,

Hproducts > Hreactants

H is positive (endothermic)

Calculating Heat of Reaction

How much heat will be released when 34.0g of hydrogen peroxide decomposes?

2 H2O2(l) 2 H2O(l) + O2(g) ΔH° = -190 kJ

-190 kJ

2 mol H2O2

34.0 g H2O2 1 mol H2O2 - 190 kJ

34.0 g H2O2 2 mol H2O2

= - 95.0 kJ

Draw an energy diagram for this reaction.

Ratio of coefficients

2. How much heat will be released when 184 g of NO2 is formed at STP?

N2(g) + 2 O2(g) 2 NO2(g) ΔH° = +68.0 kJ

184 g NO2 1 mol NO2 68.0 kJ

46 g NO2 2 mol NO2

= 136 kJ

Draw an energy diagram for this reaction.

1. What type of

reaction is this?

2. Would the value of ΔH be positive

or negative?

1. What type of

reaction is this?

2. Would the value of ΔH be positive

or negative?

Changes of State

• Melting= s l ΔH= • Freezing= l s ΔH=• Condensation= g l ΔH=• Evaporation/Boiling= l g ΔH=• Deposition= g s ΔH=• Sublimation= s g ΔH=

Changes of State

• Heat of vaporization- The amount of heat required to convert a unit mass of a liquid at its boiling point into vapor without an increase in temperature.

• Heat of fusion- the amount of heat required to convert a unit mass of a solid at its melting point into a liquid without an increase in temperature.

Molar Heat of Fusion / Vaporization

• How much heat does it take to melt / evaporate something.

• Expressed in kJ / mol

Energy (kJ)

mol

Molar Heat =

Sample Problem A

How much energy is absorbed when 2.61 mol of ice melts? (Molar Heat of Fusion for ice = 6.009 kJ / mol)

Step 1:Outline what you know.Molar Heat = 6.009 kJ / molEnergy = ? kJmol = 2.61 mol

Sample Problem A

How much energy is absorbed when 2.61 mol of ice melts? (Molar Heat of Fusion for ice = 6.009 kJ / mol)

Step 2:Plug into the equation.

Energy (kJ)

2. 61 mol

6.009 =

Sample Problem A

How much energy is absorbed when 2.61 mol of ice melts? (Molar Heat of Fusion for ice = 6.009 kJ / mol)

Step 3:Solve. Energy

(kJ)

2. 61 mol

6.009 =(2.61 mol)(2.61

mol)

Sample Problem A

How much energy is absorbed when 2.61 mol of ice melts? (Molar Heat of Fusion for ice = 6.009 kJ / mol)

Step 3:Solve.

Energy = 15.7 kJ

Changes of State

QUIZ 12-12

2 Fe + 1.5 O2 Fe2O3

∆H= -48.0kJ

How much heat is released when 112 g of Fe reacts?

QUIZ 12-14

How much heat would be produced by burning 44.8 L of oxygen at STP?

C2H5OH (l)+3O2(g)2CO2 (g)+3 H2O(l)

∆H = -930 kJ