Download Thermodynamics–Honors

 Ability to do work
 Units– Joules (J), we will use “kJ”
 Can be converted to different types
 Energy change results from forming and breaking
chemical bonds in reactions
1) Kinetic Energy– energy of “motion”
2) Potential Energy– “stored” energy
1) Open System
1) Closed System
1) Isolated System
 Energy transfer between a system and the surroundings
 Transfer is instant from high----low temperature until
equilibrium
 Temperature—
 Measure of heat, “hot/cold”
 the average kinetic energy of molecules
 Kinetic theory of heat
 Heat increase resulting in temperature change causes an
increase in the average motion of particles within the
system.
 Increase in heat results in
 Energy transfer
 Increase in both potential and kinetic energies
 First Law of Thermodynamics
 Energy is conserved in a reaction (it cannot be created or
destroyed)---sound familiar???
 Math representation: ΔEtotal = ΔEsys + ΔEsurr = 0
 Δ= “change in”
 ΔΕ= positive (+), energy gained by system
 ΔΕ= negative (-), energy lost by system
 Total energy = sum of the energy of each part in a chemical
reaction
 Heat = transfer of energy
 Temperature = measurement of heat
 System = the area or space we focus on
 Surroundings = everything else apart from the system
 Boundary = separates system and surroundings
How do we find the change in energy/heat transfer that
occurs in chemical reactions???
 Experimentally “measuring” heat transfer for a chemical
reaction or chemical compound
 Calorimeter
 Instrument used to determine the heat transfer of a chemical
reaction
 Determines how much energy is in food
 Observing temperature change within water around a
reaction container
** assume a closed system, isolated container
 No matter, no heat/energy lost
 Constant volume
 Amount of heat required to increase the temperature of 1g
of a chemical substance by 1°C
 Units: cal/g-K or J/g-K
 4.184 J = 1 cal,
K = 273 + °C
 Allows us to calculate how much heat is released or
absorbed by a substance ! ! !
 Unique to each chemical substance
 Al(s) = 0.901J/g°K
 H2O(l) = 4.18 J/g°K
 q = smΔΤ
 s/Cp = specific heat (values found in reference table)
 m = mass in grams
 ΔΤ= change in temperature
Q=?
m = 420 g
C(H2O (l)) = 4.18 J/g• C
ΔT = 37-25 = 12 C
Q = mc ΔT
Q = (420 g)(4.18 J/g•  C)(12 C)
Q = 21067 J or 21 kJ
Q=?
m = 755 g
cFe = 0.45 J/g•C
ΔT= 132 C - 12 C = 120 C
Q = mc ΔT
Q = (755g)(0.45 J/g•C)(120 C)
Q = 40,700 J
 Styrofoam cup with known water mass in calorimeter
 Assume no heat loss on walls
 Initial water temp and then chemical placed inside
 Final temperature recorded
 Any temperature increase has to be from the heat lost
by the substance SOOO
 All the heat lost from the chemical reaction or substance
is transferred to H2O in calorimeter
qchemical = -qwater
 The specific heat of gold is 0.128 J/g°C. How much
heat would be needed to warm 250.0 g of gold from
25°C to 100°C?
 Calorimetry Worksheet
 Fusion means melting/freezing
 Vaporization means boiling/condensing
 Hf and Hv - amount of energy needed to
melt/freeze or boil/condense 1g of a substance
 Different for every substance – look on reference
tables
 Q = mHf
 Q = mHv
 Calculate the mass of water that can be frozen by
releasing 49370 J.
 Calculate the heat required to boil 8.65 g of
alcohol (Hv = 855 J/g).
 Calculate the heat needed to raise the
temperature of 100. g of water from 25 C to 63 C
.
 The flat points represent a phase change – temperate
does not change while a phase change is occurring
even though heat is being added.
 Diagonal points represent the 3 phases