Intro to Thermodynamics: Heat and Temperature

Temperature As objects get warmer, the particles within them move faster. The kinetic energy of these particles is called Internal Energy (U)

Temperature Temperature measures average internal energy of an object We sense temperature as “warm” or “cold”

Which has more thermal energy? Gallon of water at 15ºC Teaspoon of water at 20ºC Temperature does not tell you the AMOUNT of thermal energy in a substance

Thermal Energy (Heat) Depends on: Mass mass T.E. 2. Temp temp T.E. 3. Substance Different substances store different amounts of energy (we will learn more about this later in the unit!)

Temperature Can the object get any colder? If an object cools, the particles will move more slowly Eventually, the particles will all come to a complete stop Can the object get any colder? This temperature is known as ABSOLUTE ZERO At absolute zero the internal energy is zero!!

Temperature Scales 3 major scales to measure temperatures: Fahrenheit: oldest; used in US Celsius (centigrade): used in most of world Based on water properties freezing point (0oC) & boiling point (100oC) Kelvin: used in sciences Based on absolute zero

Temperature Examples No º sign oF oC K Boiling Water 212 100 373 Human Body Temp. 98.6 37 310 Room Temp 70 21 294 Freezing Water 32 0 273 Absolute Zero -460 -273 0

Conversions oC = (5/9)(oF - 32) oF = (9/5)(oC) + 32 K = oC + 273

Conversion Example 100 degrees Fahrenheit to Celsius oC = 37.8o oC = (5/9)(oF - 32) oC = (5/9)(100 - 32) oC = (5/9)(68) oC = 37.8o

Example ºF ºC K -72º F 156ºC 35 K -72oF to oC oC = -57.8o 215.2 -72oF to oC oC = (5/9)(oF - 32) oC = (5/9)(-72 - 32) oC = (5/9)(-104) oC = -57.8o -57.8oC to Kelvin K = oC + 273 K = -57.8 + 273 K = 215.2

Example ºF ºC K -72º F 156ºC 35 K 156°C to °F oF = 312.8o -57.8 215.2 312.8 429 156°C to °F oF = (9/5)(oC) + 32 oF = (9/5)(156) + 32 oF = 280.8 + 32 oF = 312.8o 156°C to Kelvin K = oC + 273 K = 156 + 273 K = 429

Example ºF ºC K -72º F 156ºC 35 K Kelvin to°C oC = -238o -238°C to °F -57.8 215.2 312.8 429 -396.4 -238 Kelvin to°C K = oC + 273 35 = oC + 273 oC = -238o -238°C to °F oF = (9/5)(oC) + 32 oF = (9/5)(-238) + 32 oF = -428.4 + 32 oF = -396.4o

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Thermo Day 2 2/13

What happens when objects with different temps come in contact? Ice cubes at 0ºC Tea at 45ºC

What happens when objects with different temps come in contact? Ice cubes at 0ºC Tea at 45ºC What happens to the temperature of the ice cubes? What happens to the temperature of the tea? When do these temperatures stop changing?

Thermal Equilibrium Occurs when two objects in physical contact reach the SAME TEMPERATURE The new temperature lies somewhere between initial temperatures

Thermal Energy (Heat) Total Internal Energy of an Object Can transfer to other objects because of a difference in temperatures **always flows from HOT to COLD** Present in all objects—EVEN COLD ONES!!

Transfer of Thermal Energy 3 basic methods: Conduction 2. Convection 3. Radiation

Transfer of Thermal Energy Convection: The transfer of heat by the actual motion of a fluid (liquid or gas) in the form of currents Conduction: The transfer of heat by direct contact of particles of matter Radiation: Heat transfer by electromagnetic waves

**Only transfer method that can travel through a vacuum— Radiation heat transfer by electromagnetic waves **Only transfer method that can travel through a vacuum— no MATTER necessary** Ex. SUNLIGHT Features: Ultraviolet Visible Light Color spectrum – (ROYGBIV) Infrared

Conductors --allow heat to transfer easily Ex: Metals Insulators --do not allow heat to transfer easily Ex: Wood, Gases

Reflectors vs Absorber Reflector: Substance that turns back radiation before it is absorbed ex: metals, white colors Absorber: Converts RADIANT energy into THERMAL energy ex: solar cells, Dark colors

Emitters Objects that give off radiation; transfer energy to their surroundings **Objects that are good absorbers are good emitters Ex: Black top roads

Heat Transfer within a System System—interactions between objects isolated for study; does not include environment **Heat always flows from HOT to COLD Energy transfers from one object to another because of a temperature difference!!

Thermo Day 3 2/14

Something to think about as you come in! The Law of Conservation of Energy states that energy will never be created or destroyed. It can only be transferred from one form to another Do you think you could transfer energy from one object to another within the system? Will the energy still be conserved?

Heat Transfer within a System ENERGY LOST BY ONE OBJECT IS EQUAL TO THE ENERGY GAINED BY THE OTHER This applies to thermal energy (heat)!! The change in Thermal Energy is represented by a Q and will be measured in Joules or calories Q gained by object 1 = Q lost by object 2

Measurements The calorie Measure of heat energy required to raise temperature of 1 gram of water by 1°C 1 calorie = 4.186 Joule

The substance’s ability to absorb heat energy Heating substances Different substances experience different temperature changes when the same amount of heat energy is added to them! This is because of the… Specific Heat The substance’s ability to absorb heat energy Every substance has its own Specific Heat! It’s like a finger print and is can be used to identify a substance!

Specific Heat Cwater = 1 calorie/goC = 4.186 J/goC Symbol = c Units = calorie/goC OR J/goC Examples: Water—VERY HIGH SPECIFIC HEAT slow to heat up (requires lots of energy) slow to cool down. Cwater = 1 calorie/goC = 4.186 J/goC Metals –low specific heats heat up and cool off quickly

Specific Heat of various substances Air Aluminum Copper Glass Ice (-20 to 0 0 C) Iron Mercury Ocean Water Water Wood 0.25 0.22 0.09 0.20 0.50 0.11 0.03 1.00 0.42 0.93 (calorie/goC) Specific Heat of various substances

Heat Transfer within a System Remember this?? Thermal Energy Depends on: Mass mass T.E. 2. Temp temp T.E. 3. Substance

Q = mc T HEAT EQUATION Q = Heat energy in Joules or calories m = Mass in grams (or kilograms) c = Specific Heat in calories/goC or J/goC (or calories/kgoC or J/kgoC) either given or found in a chart T = Change in Temperature **(high temperature - low temperature) = oC

Calculating Heat Example How much heat is needed to raise the temperature of 250 grams of water from 20°C to 45°C? (cw = 4.186 J/goC) Given Equation Plug-n-Chug Answer Q = ? Q = mc T m = 250 g Th = 45oC Q = (250)(4.186)(45-20) Tc = 20oC Q = 26162.5 J cw = 4.186 J/goC

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Thermo Day 5 2/16

Q = mc T HEAT EQUATION Q = Heat energy in Joules or calories m = Mass in grams (or kilograms) c = Specific Heat in calories/goC or J/goC (or calories/kgoC or J/kgoC) either given or found in a chart T = Change in Temperature **(high temperature - low temperature) = oC

Calculating Heat Example A 0.050kg metal bolt is heated to 81°C. It is then dropped into a beaker containing 0.15 kg of water with an initial temperature of 25°C. If the metal has a specific heat capacity of 899 J/kg°C, what will be the final temperature of the bolt and water?

Law of Conservation of Energy- Remember this slide?? Law of Conservation of Energy- ENERGY LOST BY ONE OBJECT IS EQUAL TO THE ENERGY GAINED BY THE OTHER Q gained by object 1 = Q lost by object 2

Calculating Heat Transfer A 0.050kg metal bolt is heated to 81°C. It is then dropped into a beaker containing 0.15 kg of water with an initial temperature of 25°C. If the metal has a specific heat capacity of 899 J/kg°C, what will be the final temperature of the bolt and water? (cw = 4186 J/kgºC) Given Equation Plug-n-Chug Answer T2bw = ? Qw = Qb mb = 0.050 kg mc T = mc T T1b = 81oC (0.15)(4186)(T2bw- 25) = (0.050)(899)(81-T2bw) mw = 0.15 kg (627.9)(T2bw- 25) = (44.95)(81-T2bw) T1w = 25oC 627.9T2bw- 15697.5 = 3640.95 - 44.95T2bw cb = 899 J/kgºC T2bw = 28.74oC 672.85T2bw = 19338.45 cw = 4186 J/kgºC

Conservation of Energy 1st Law of Thermodynamics-- Conservation of Energy Change in Internal Energy Energy can not be created or destroyed, but can only change form ∆Ep + ∆Ek + ∆U = 0 Energy can be transferred between objects as either : WORK or HEAT

HEAT Internal Energy vs Kinetic Energy Energy transferred of particles within object Energy transferred to or from object

Changing the Internal Energy There are two ways to change the internal energy of a system: Wby = -Won WORK Thermal reservoir a) by system on environment Internal energy decreases b) on system by environment Internal energy increases HEAT a) from system to environment Internal energy decreases b) to system from environment Internal energy increases

Heat Engines Remember PV = nRT ? Using heat to do work Remember PV = nRT ? Raising temperature can increase volume or pressure Increasing volume does work on the environment! Gas heats and expands Thermal reservoir Thermal reservoir Piston pushes outward on environment

ENTROPY– The 2nd and 3rd Laws of Thermodynamics When left alone, everything tends toward disorder Entropy--measure of amount of disorder in a system particle kinetic energy entropy

LET’S MOVE AROUND!! Entropy Entropy of Gas > E. of Liquid > E. of Solid LET’S MOVE AROUND!!

Second Law of Thermodynamics An isolated system will spontaneously move to a state of thermal equilibrium Less Energy More Energy Less Entropy More Entropy COLD HOT Energy moves both directions More moves from hot to cold than the reverse

Second Law of Thermodynamics Thermal Equilibrium Entropy is same throughout system Energy moves each direction with equal probability

Second Law of Thermodynamics Machines cannot transform all energy to work! cannot be 100 % efficient Force applied evenly to all sides All force applied to piston Real Ideal Heat Engine “Wasted” Energy More disorder = less available energy