Download Temperature and Heat

Temperature and Heat
Energy: capacity to do work
Kinetic energy (associated with motion)
• Thermal, atoms, molecules, and ions are in motion – all matter has thermal
energy
• Mechanical, moving tennis ball, automobiles
• Electrical, e − moving through a conductor
• Sound, corresponds to compression and expansion of the space between
molecules
Potential energy (results from object’s position)
• Gravitational, possessed by a ball held above floor
• Electrostatic, associated with the separation of two dissimilar electrical charges
• Chemical, burning coal ⇒ heat ⇒ work
Thermodynamics: part of chemistry that deals with the energy changes involved in
chemical rxns and changes in the physical state of substances.
Thermochemistry: branch of Thermodynamics that deals with the evolution or
absorption of energy as heat in chemical processes
Work: w = F ⋅ d (recall F = ma) and the units of work are: kg m s −2 m = kg m2 s −2
Units of energy: Joule (J): kg m2 s −2
Calorie (cal): energy required to raise the temperature of 1.0 g of water by 1oC (from
14.5 to 15.5 oC)
1 cal = 4.184 J and 1 J = 0.2390 cal
1 kJ = 1000 J
Nutritional calorie (Cal): energy available from food (1 Cal = 1000 cal)
• The temperature of an object is a measure of its heat content and of its ability to
transfer heat. We measure temperature by thermometers.
HEAT IS NOT THE SAME AS TEMPERATURE!
The more thermal energy a substance has, the greater the motion of atoms and
molecules.
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System: object or collection of objects under study
Surroundings: everything outside the system can exchange energy with the system
Thermal equilibrium: associated with directionality of heat transfer
Calorimetry: measures flow of heat across boundaries
Heat Capacity: measures the ability of the substance to pick up heat
Heat capacity under constant pressure, Cp equals the amount of heat required to raise the
q
temperature of a system by 1oC (1 K) under constant pressure C P =
(J K-1)
∆T
CP
q
=
⇒ q = nc p ∆T (units: J K-1 mol-1)
Molar hat capacity: cP =
n
n∆T
Heat capacity under constant volume, Cv
cP
(J K-1 kg-1) q = mcs ∆T ,
m
so cP = Mcs the molar heat capacity equals its specific heat capacity multiplied by its
molar mass.
Specific heat capacity: cs =
Heat Transfer
The quantity of heat transferred from or to an object depends on
a) the quantity of material
b) the size of temperature change
c) the identity of the material gaining or losing heat
See Table 10.1
How to find the temperature at thermal equilibrium
amt of heat gained by the cooler body = amt of heat lost by the warmer body
∆ means: final - initial
∆T = + q = + heat transferred from surroundings to system
∆T = - q = - heat transferred from system to surroundings
The heat content changes within a given system is zero.
q1 + q2 + q3 + .... = 0
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Example
55.0 g of a hot metal at 98.8oC is immersed in a beaker containing 225.0 g of cool water
at 21.0oC. The final temperature is 23.1oC. Find the specific heat capacity of the metal
and using Table 10.1 identify the metal.
qwater + qmetal = 0 assuming no loss in the surroundings
cs , water mH 2O (T f − Ti ) + cs , metal mmetal (T f − Ti ) = 0
(4.184 J K-1 g-1)(225.0 g)(296.3 – 294.2)K + cs ,metal (55.0 g)(296.3 – 371.95)K
-1 -1
c s ,metal =0.469 J K g
Fe is the closest in Table 10.1
Example
Calculate the minimum amt of heat required to raise the temperature of 45 kg of granite
by 15.0 K
The heat absorbed by the granite is:
q = mcs ∆T = ( 45 x103 g )(0.82 JK −1 g −1 )(15.0 K ) = 5.5 x105 J = 550kJ
This is the minimum heat lost to surroundings as the rock heats up
Now,
Calculate the temperature change of 45.0 kg of water if it stores the same amt of thermal
energy as the 45 kg of granite
q
5.5 x105 J
∆T =
=
= 2.9 K
cs m ( 4.184 JK −1 g −1 )( 45 x103 g )
Calorimeters
Containers that have an interior space thermally insulated from the surroundings and a
means to measure the temperature in the interior. The idea is to measure the temperature
at thermal equilibrium among well-defined systems on the inside while preventing heat
flow to or from outside
q = CCAL ∆T
part absorbs
CCAL is the calorimeter constant and it tells us how much heat the interior
Combustion (bomb) calorimeter confines the rxn into a fixed volume
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Example
A stryrofoam cup calorimeter is filled with 150.0
g of water at room temperature. Adding 1312 J
of heat to the contents raises the interior
temperature by 1.93 K. Calculate the calorimeter
constant. Assume that the calorimeter does not
leak heat.
qwater + qCAL = 1312 J
cs , water mH 2O ∆T + CCAL ∆T = 1312 J
(4.184JK -1g -1)(150.0g)(1.93K)+CCAL(1.93K)=1312 J, and CCAL =52.2 JK-1
Now we dissolve some LiCl in the 150.0 g of water contained in the same calorimeter
and we observe a temperature rise by 3.46 K. Calculate the amt of heat evolved in the
dissolution of LiCl if the heat capacity and mass of LiCl are the same as of water.
qsolution + qCAL + qrxn = 0
msolutioncs ,solution ∆T + CCAL ∆T + qrxn = 0
(150.0 g)(4.184 JK-1g-1)(3.46K) + (52.2 JK-1)(3.46K) + qrxn = 0
qrxn = - 2352 J, the dissolution rxn absorbs - 2352 J and the heat that evolves is the
opposite of it, the reverse of absorbing it,
+ 2352 J
EXAMPLE
A manufacturer claims that its new
dietetic dessert has fewer than 10 Calories
per serving. To test the claim a chemist at
the Department of Consumer Affairs
places one serving in a bomb calorimeter
and burns it in oxygen. The heat capacity
of the calorimeter is 8.151 kJ K-1. The
temperature increases 4.937oC. Is the
manufacturer’s claim correct?
qsample + qCAL = 0 ⇒ qCAL = − qsample
qCAL = (8.151 kJ K-1)(4.937K)=40.24 kJ
The heat released upon burning is gained
by the calorimeter
10 Calories = (10 kcal)(4.184 kJ/kcal) = 41.84 kJ because 1 Calorie = 1 kcal
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And 41.84 kJ > 40.24 kJ and the manufacturer is honest!!!
EXAMPLE
A chemist burns 0.865 g. of graphite in a new bomb calorimeter and CO2 is formed. If T
increases 2.613 K and 393.5 kJ of heat is released per mole of graphite, what is the heat
capacity of the calorimeter?
qsample + qCAL = 0
(0.865g )(
1molC
)( −393.5kJmol −1 ) + (2.613K )CCAL = 0
12.0 gC
CCAL = 10.85kJK −1
Enthalpy
The heat content of a substance under constant pressure, ∆H = qP. It is a state property.
∆H= Hf - Hi
Enthalpy of a rxn is the change in enthalpy when a chemical rxn occurs
∆H= Hproducts – Hreactants
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) ∆Hrxn = qp < 0 exothermic
N2 (g) + O2 (g) → 2 NO2 (g) ∆Hrxn = qp > 0 endothermic
Enthalpies of rxn are written in association with a balanced equation
CO (g) + ½ O2 (g) → CO2 (g) ∆Hrxn = - 283 kJ mol-1
2 CO (g) + O2 (g) → 2 CO2 (g) ∆Hrxn = - 566 kJ mol-1
EXAMPLE
The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its
thermal decomposition can be represented by
Al2O3 (s) → 2Al (s) + 3/2 O2 (g) ∆H rxn = 1676 kJ. How many grams of Al can be
formed when 1.000 x 103 kJ of heat is transferred?
(1.000 x 103 kJ) x (2 mol Al/1676 kJ)(26.98 g Al/1 mol Al) = 32.20 g Al
Changes in State (Phase diagrams)
Common phase transitions involve a transfer of heat between system and surroundings
at constant temperature and pressure.
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GAS
Condensation
Vaporization
Sublimation
Deposition
LIQUID
Fusion to m.p.
Freezing
SOLID
EXAMPLE
Heating of 500.0 g of water from -50oC to 200oC (Heat of fusion: 333 J/g, heat of
vaporization: 2256 J/g). How much is needed for this process to occur?
Step 1: warm water from -50oC to 0oC (cs of ice = 2.06 J K-1 g-1)
q1 = cs,ice mice ∆T = (2.06 J K-1 g-1)(500.0 g)(273.15 – 223.15)K = 5.15 x 104 J
Step 2: melt ice at 0oC
q2 = (500.0 g)(333 J g-1) = 1.67 x 105 J
Step 3: provide heat to raise the temperature of water from 0oC to 100oC
q3 = cs , water mH 2O ∆T = (4.184 J K-1 g-1)(500.0 g)(373.15-273.15)K = 1.13 x 105 J
Step 4: provide heat to evaporate water at 100oC
q4 = (500.0 g)(2256 J g-1) = 1.13 x 106 J
Step 5: raise the temperature of steam from 100oC to 200oC
q5 = cs , steammwater ∆T = (1.92 J g-1)(500.0 g)(473.15-373.15)K = 9.60 x 104 J
qTOTAL = q1 + q2 + q3 + q4 + q5 = 1.60 x 106 J = 1600 kJ
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Determining the heat of a reaction
You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.00oC and
carefully add 25.0 mL of 0.500 M HCl, also at 25.00oC. After stirring, the final
temperature is 27.21oC. Calculate qsoln (in J) and ∆Hrxn (in kJ/mol). (Assume the total
volume is the sum of the individual volumes and that the final solution has the same
density and specific heat capacity as water: d = 1.00 g/mL and c = 4.184 J/g K)
1. We find the mass of the solution
msoln = (25.0 mL + 50.0 mL)(1.00 g/mL) = 75.0 g
2. We find ∆Tsoln, ∆Tsoln = 27.21oC - 25.00oC = 2.21 oC = 2.21K
3. Now we find qsoln
qso ln = mso ln cs , so ln ∆Tso ln = (75.0 g )(4.184 JK −1 g −1 )(2.21K ) = 693J
4. Now we find ∆Hrxn
We have a neutralization rxn HCl + NaOH → NaCl + H2O
Let us write the net ionic equation: H+ (aq) + OH − (aq) → H2O (l)
We find the moles of reactants and products
n H + = (0.500mol / L)(0.0250L) = 0.0125
nOH − = (0.500mol / L)(0.050L) = 0.0250
∴ H+ is the limiting reactant and only 0.0125 mol of H2O will be formed.
The heat gained by water was lost from the rxn:
qsoln = - qrxn ⇒ qrxn = - 693 J
∆H rxn =
qrxn
1kJ
− 693J
1kJ
x
=
x
= −55.4kJmol −1
molH 2O 1000 J 0.0125 1000 J
↑(to get the units requested)
ENTHALPY CHANGES ACCOMPANY CHEMICAL REACTIONS
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Summary of the relationship between amount (mol) of substance and the heat (kJ)
transferred during a reaction.
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