energy and temperature
DESCRIPTION
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 - PowerPoint PPT PresentationTRANSCRIPT
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