Thermodynamics

Energy 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

Basic Energy Types 1)Kinetic Energy– energy of “motion”

2)Potential Energy– “stored” energy

System vs. Surroundings

Types of Systems

1)Open System

2)Closed System 3)Isolated System

Heat (q)

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

Heat (q) continued 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 inEnergy transferIncrease in both potential and kinetic energies

Thermodynamics 101First Law of Thermodynamics

Energy is conserved in a reaction (it cannot be created or destroyed)---sound familiar???

Math representation: ΔEtotal = ΔEsys + ΔEsurr = 0Δ= “change in” ΔΕ= positive (+), energy gained by systemΔΕ= negative (-), energy lost by system Total energy = sum of the energy of each part in a

chemical reaction

Basic TerminologyHeat = transfer of energyTemperature = measurement of heatSystem = the area or space we focus onSurroundings = everything else apart from

the systemBoundary = separates system and

surroundings

CalorimetryHow do we find the change in energy/heat

transfer that occurs in chemical reactions???

Calorimetry

Experimentally “measuring” heat transfer for a chemical reaction or chemical compound

Calorimeter Instrument used to determine the heat transfer of a

chemical reactionDetermines how much energy is in food Observing temperature change within water around a

reaction container

** assume a closed system, isolated containerNo matter, no heat/energy lost Constant volume

Specific HeatAmount 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 + °CAllows us to calculate how much heat is released or

absorbed by a substance ! ! !

Unique to each chemical substance Al(s) = 0.901J/g°KH2O(l) = 4.18 J/g°K

Specific Heat Equations

q = smΔΤs/Cp = specific heat (values found in

reference table)m = mass in gramsΔΤ= change in temperature

Example 1: How much energy is required to warm 420 g of water in a water bottle from 25C to 37C ?

Q = ? m = 420 gC(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

Example 2: How much energy is released

by cooling 755 g of iron from 132 C to

12 C ?

Q = ? m = 755 g cFe = 0.45 J/g•C

ΔT= 132 C - 12 C = 120 CQ = mc ΔT

Q = (755g)(0.45 J/g•C)(120 C)Q = 40,700 J

“Coffee Cup” calorimeterStyrofoam cup with known water mass in

calorimeterAssume no heat loss on wallsInitial water temp and then chemical placed

insideFinal temperature recorded

Any temperature increase has to be from the heat lost by the substance SOOOAll the heat lost from the chemical reaction or

substance is transferred to H2O in calorimeter

“Coffee Cup” calorimeter (cont.)

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?

Example 3:

Calorimetry Worksheet

Homework

Heat of Fusion (Hf) /Heat of Vaporization(Hv)Fusion means melting/freezingVaporization means boiling/condensingHf 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

Examples: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 .

Phase Change Diagram

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