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Week 5/Tu: Thermo, Lecture Unit ‘11’ © DJMorrissey, 2o12 Exam 1: elements, compounds, mass, and reactions Unit 11: Thermodynamics -- definitions, heat, work, system, ΔE -- 1 st law -- energy and temperature -- specific heat vs. heat capacity Unit 12: Heat capacity -- heating curves, change of state, -- “PV” work, enthalpy Issues: ? No Homework this week http://www.quickmeme.com/meme/5mbj/
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Transcript of Week 5/Tu: Thermo, Lecture Unit ‘11’ · Week 5/Tu: Thermo, Lecture Unit ‘11’ ©...

• Week 5/Tu: Thermo, Lecture Unit 11

DJMorrissey, 2o12

Exam 1: elements, compounds, mass, and reactions Unit 11: Thermodynamics -- definitions, heat, work, system, E -- 1st law -- energy and temperature -- specific heat vs. heat capacity Unit 12: Heat capacity -- heating curves, change of state, -- PV work, enthalpy Issues: ? No Homework this week

http://www.quickmeme.com/meme/5mbj/

• Week 5/Tu: Electron Transfer, redo

DJMorrissey, 2o12

Retry the demonstration: 2 Mg (s) + CO2 (g) 2 MgO (s) + C (s) element + oxide metal oxide + element Oxidation numbers: Mg 0 Ox -2, C +4 Ox -2, Mg +2 C 0 oxidation (increase) 2* [ Mg 0 +2 ] reduction (decrease) C +4 0 Notice that the changes must balance (electrons are moving between atoms and we cant create or destroy electrons)

• Week 5/Tu: Thermodynamics, the system

DJMorrissey, 2o12

Thermodynamics started literally as the study of heat flow and work production in physical systems (e.g., steam powered machines) but evolved into the much more general study of energy and work flow in any physical system. The basic finding was that all forms of energy and work are equivalent and can be interconverted (in principle) but there are fundamental limits on the efficiency of conversion. The critical starting point for any thermodynamic analysis is to clearly define the system and the surroundings. Speaking colloquially, changes in the energy and the work done can only be determined if the system is complete. We need to be able to draw a boundary around the system through which only energy or work can pass. This sounds simple but many erroneous applications of thermodynamics involve incorrect definitions of the system boundaries. Similarly, it is very difficult in practice to construct such boundaries. Thermodynamics provides a theoretical reference to judge real systems.

Recall Demo: Sugar + H2SO4 Easy to define system?

• Week 5/Tu: Chemical Thermodynamics

DJMorrissey, 2o12

Focusing on chemical systems: the energy transfer should be easy to measure but the work may not be so apparent. Energy types to keep track of: Light photochemistry, chemiluminescence Electrical electrochemistry, RedOx reactions Heat thermochemistry (endothermic v. exothermic reactions) (internal) Energy kinetic energy of motion (K.E. gas > K.E. liquid) potential energy from from forces (gravity, electrical) We have seen evidence of internal chemical potential energy Chemical potential energy is due to the difference in the electrical force felt by an electron on one atom when it encounters the electrical field of a different atom plus the internal kinetic energy of motion. We will have to develop a way to keep track of the chemical energy.

• Week 5/Tu: Chemical Thermodynamics

DJMorrissey, 2o12

Focusing on chemical systems: the energy transfer should be easy to measure but the work may not be so apparent. Work types include: Electrical moving current in external circuit (battery) Expansion Pressure * Volume work (easy to overlook)

• Week 5/Tu: First Law

DJMorrissey, 2o12

Focusing on a chemical process in a properly defined chemical system: the change in the internal energy of that system during a (chemical) process must equal the amount of heat and the amount of work that cross the system boundary.

Conservation of energy: (Eafter Ebefore) = E = q + w

E is the internal energy q is the heat transfer w is the work transfer

The division between heat and work in thermodynamics is largely historical since the field initially focused on steam engines and the like. Direction of energy flow or sign convention: positive values of q and w correspond to increasing the internal energy of the system. Thus, if a system delivers heat to the outside surroundings, the sign of q must be negative.

[Later in the course we will consider the Second Law]

• Week 5/Tu: Temperature and Energy

DJMorrissey, 2o12

What is temperature? Temperature is an intensive variable that represents the internal energy of a substance. (Remember that heat is simply a form of energy.) Where does the energy (heat) go or what does it do when an object is heated? The energy is absorbed into internal motion(s) of the atoms or molecules. What are the SI dimensions (units) of energy? (James) Joule = force * distance

force = mass * acceleration (Isaac) Newton acceleration = velocity / time velocity = distance / time Joule = (mass * (distance/time)/time ) * distance = kg * m2 /s2

What is power? Power is energy / time, Joule/s = (James) Watt

• Week 5/Tu: Simple Energy Question

DJMorrissey, 2oo9

Q: Consumers Power (a local energy company) said that they delivered 293 kW-hours to my house in August and sent me a bill. They claim they delivered energy, so how many Joules is this?

# Joules = 293 kW-hr * (103 W/kW) * (1 J-s /W) * (3600 s/hr) = 1.05x109 J = 1.05 GJ

• Week 5/Tu: Temperature and Specific Heat

DJMorrissey, 2o12

I assume that everyone is familiar with the concept that some things heat up quickly and others slowly. Similarly, for cooling Why is that? DATA: measure the temperature change after adding heat (energy) for a 1kg object.

kJ/kg/K kJ/kg/K Acetic Acid (l) 2.04 Cadmium (s) 0.06 Bismuth (s) 0.122 Bismuth (s) 0.122 Bismuth (l) 0.15 Mercury (l) 0.14 Bromine (l) 0.47 Bismuth (l) 0.15 Cadmium (s) 0.06 Diamond (s) 0.15 Copper (s) 0.385 NaCl (s) 0.21 Diamond (s) 0.15 Copper (s) 0.385 Mercury (l) 0.14 Bromine (l) 0.47 NaCl (s) 0.21 Sulfuric Acid (l) 1.34 Sulfuric Acid (l) 1.34 Water (g) 1.9 Water (l) 4.19 Acetic Acid (l) 2.04 Water (g) 1.9 Water (l) 4.19

Alphabetical Order Specific Heat Order

• Specific heat kJ/kg-K Molar Mass kJ/mol-K

(molar) Heat Capacity kJ/mol-K

Cadmium (s) 0.06 0.112 0.54 Cadmium (s) 0.54 Bismuth (s) 0.12 0.209 0.58 Bismuth (s) 0.58 Mercury (l) 0.14 0.201 0.70 Mercury (l) 0.70 Bismuth (l) 0.15 0.209 0.72 Bismuth (l) 0.72 Diamond (s) 0.15 0.012 12.5 NaCl (s) 3.59 NaCl (s) 0.21 0.059 3.59 Bromine (l) 5.88 Copper (s) 0.385 0.063 6.06 Copper (s) 6.06 Bromine (l) 0.47 0.0799 5.9 Diamond (s) 12.50 Sulfuric Acid (l) 1.34 0.098 3.47 Sulfuric Acid (l) 13.66 Water (g) 1.9 0.018 106. Acetic Acid (l) 34.00 Acetic Acid (l) 2.04 0.060 34. Water (g) 105.56 Water (l) 4.19 0.018 233. Water (l) 232.78

Week 5/Tu: Specific Heat vs. Heat Capacity

DJMorrissey, 2o12

I assume that everyone is familiar with the concept that some things heat up quickly and others slowly. Similarly, for cooling Why is that? DATA: recall that we should be thinking about atoms & molecules, not masses.

Specific Heat / Molar Mass = Heat Capacity [ be careful about units]

• Week 5/Tu: Conservation of Energy Question

DJMorrissey, 2o12

Some people say that you should put a metal spoon in a cup before adding hot water to make tea, avoid breaking. Q: What is the final temperature for 250 mL of water at 95o C if an iron spoon (25 g) at 25oC is put into the water? HINT: Conservation of energy Energy lost cooling water = energy gain heating spoon Energy loss (water)= 250mL*1.0 g/cm3 *4.19 J/g-K *(Tf-95) = 1048 J/K *(Tf-95) Energy gain (spoon)= 25g*0.45 J/g-K *(Tf-25)

= 11.25 J/K *(Tf-25) 1048 J/K *(Tf-95) +11.25 J/K *(Tf-25) = 0 Tf = 93.7