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 Kinetic Energy– energy of “motion” Potential Energy– “stored” energy

System vs. Surroundings

Types of Systems Open System Closed System 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 Increase in heat results in 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

Thermodynamics 101 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

Basic Terminology 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

Calorimetry How 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 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

Specific Heat 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

Specific Heat Equations q = smΔΤ s/Cp = specific heat (values found in reference table) m = mass in grams ΔΤ= change in temperature

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

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

“Coffee Cup” calorimeter 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

“Coffee Cup” calorimeter (cont.) qchemical = -qwater

Example 3: 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? 2400J

Homework Calorimetry Worksheet

Heat of Fusion (Hf) /Heat of Vaporization(Hv) 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

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