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THERMODYNAMICS Lecture 1 - Energy, Heat, and Work - The First Law of Thermodynamics - The Second Law of Thermodynamics - Thermal Processes. 1) What is the definition of THERMODYNAMICS? 2) What are the laws of THERMODYNAMICS? 3) Define system, boundary, and surroundings. Do Now.
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THERMODYNAMICSLecture 1- Energy, Heat, and Work - The First Law of Thermodynamics - The Second Law of Thermodynamics - Thermal Processes
1) What is the definition of THERMODYNAMICS? 2) What are the laws of THERMODYNAMICS? 3) Define system, boundary, and surroundings. Do Now
THERMO DYNAMICS heat movement Thermodynamics is the study of the effects of work, heat, and energy on a system. Do Now Solutions
Thermodynamics is concerned with: - Forms of energy - Transformation of energy - The direction of natural process Energy
First law of thermodynamics – change in internal energy of a system is the difference between the heat taken in by the system and the work done on the system ∆U= Q + W. Do Now Solutions
Second law of thermodynamics –we can’t convert all the heat energy into work. Zeroth law of Thermodynamics – states that two objects, each in thermal equilibrium with a third object, are in thermal equilibrium with each other. Do Now Solutions
-- Concept Check -- If we want to observe the temperature of water in a tea kettle as it sits on the stove, what would be our system, boundary, and surroundings? Do Now Solutions • System – the part of the universe chosen for study • Surroundings – everything outside of the system • Boundary – the barrier separating a system from its surroundings
<Source: id.mind.net/.../introduction/introduction.html> It is hard to visualize energy as an object. An object, let's say a basketball, may possess energy, and we can certainly visualize it. However, the basketball is not energy. Energy
Kinetic energy (KE): energy of motion KE = ½ mv² Potential energy (PE): stored or positional energy PE = mgh e.g. gravitational potential energy and a spring. Kinetic and Potential energies are applied to objects on macroscopic level – objects that you and I can see Forms of Energy
Forms of Energy Source:<http://www.qrg.northwestern.edu/projects/vss/does/space-environment/1-what-is-energy.html> • Kinetic and Potential energies are applied to objects on macroscopic level
Internal energy (U): a combination of different energies: -Kinetic -Potential -Chemical -Vibrational -Rotational Internal energy (U) is an energy on microscopic level Forms of energy Molecules?
A good definition of energy is: Energy is the ability to do work. A good definition of work is: Work is the transfer of energy. Basic Definition
What do we know so far about work? W = F × d Work
How do we go from the work learned in physics to the work used in thermodynamics? W = Fapplied∆x = -Fsystem ∆x Pressure (P) = Force / Area → P = F / A P = F / A → F = P × A W = -Fsystem ∆x = - (P × A) ∆x Since (A × ∆x) = ∆V → WEC = -P∆V Work
Work = F × d = (kg × m/s²)× m JOULES = kg × m²/s² Units For Work
A gas is confined in a cylinder by a piston. The initial volume is 0.10 m³. The piston is held in place by latches in the cylinder wall. The whole apparatus is placed in air at atmospheric pressure of 101.3 kPa. What is the work done by the system if the gas was to expand to double its initial volume? Problem
W = - F ∆x = - P ∆V = - (101.3 kPa) × (0.2-0.1) m³ = -10.13 (kN/m²) × m³ = -10.13 kN× m = -10.13 kJ Note: Since W is work done by the system, it is negative Solution
Law of Conservation of Energy: “Energy can't be created or destroyed” You can only change it from one form to another. The First Law of Thermodynamics
The First Law of Thermodynamics • Source: <www.txucorp.com/.../images/stea_tur.gif>
Q is positive when the system gains heat and negative when it loses heat. W is positive when work is done on the system and negative when work is done by the system. Sign Convention
Ideal Gas Problem 1 Air in a cylinder and piston system initially has a volume of 10 L, a pressure of 101.3 kPa, and a temperature of 25°C. What would be the final temperature of the system if it were compressed to half its volume and 1.013 MPa?
Solution P1V1 / T1 = P2V2 / T2 (101.3 kPa) (10 L) / (298 K) = (1,013 kPa) (5 L) / T2 T2 = 1490 K
Sample Problem 1 One thousand joules of mechanical work are done by an insulated system that expands when 5.5 kcal of heat are added. What is the change in the internal energy of the system? (4,187 joules/kcal)
Use the first law of thermodynamics: ∆U = Q + W ∆U = (5.50 kcal) (4187 J/kcal) + (-1000 J) = 22 kJ Solution
Gas confined by the piston in a particular heat engine expands against a constant pressure of 2.56e5 N/m2. When 40 kJ of heat are added to the system, the volume expands from 0.105 m3 to 0.235 m3. What is the change in internal energy of the system? Sample Problem 2
Use the first law of thermodynamics ∆U = Q+W = Q-P∆V ∆U = 40 kJ-(2.56 e5 N/m²(0.235 m³-0.105 m³)) ∆U = 7.7 kJ Solution
Adiabatic process? Isothermal process? Isochoric process? Isobaric process? Do Now
Adiabatic process – a process during which no heat enters or leaves the system Isothermal process – no change in the temperature of the system Isochoric process – no change in volume Isobaric process – no change in pressure Do Now Solutions
Isobaric Process: Isobaric Process is one that occurs at constant pressure Thermal Processes
Isobaric process Work Done
Isochoric process is a thermodynamic process that occurs at constant volume Isochoric Process
