Heat A Form of Energy Thermal energy 12

Heat A Form of Energy (Thermal energy)

12. 1 Temperature & Thermal Energy Objectives • Explain heat using the Kinetic-Molecular Theory • Define temperature • Understand the process of thermal equilibrium • Describe and use the Celsius & Kelvin Temperature Scales

Kinetic-Molecular Theory It was once common belief that heat was an invisible substance called “caloric. ” It was believed that it could be transferred between objects. To heat up an object this caloric had to flow into it. This, they thought, explained why objects expanded when heated. (But……could not explain heat coming from a cold log when it was burned) The Kinetic-Molecular Theory replaced the “Caloric Theory” in the 19 th century. It stated that all matter is made up of atoms/molecules in constant motion. The faster they move, the hotter an object will be.

Molecules and Motion • The motion of molecules produces heat • The more motion, the more heat is generated

Thermal Energy (also called Internal energy) is the energy an object or substance has due to the kinetic and potential energies associated with the random motions of all the particles that make it up. The hotter something is, the faster its molecules are moving/vibrating, the higher its temperature.

Temperature and Heat • Kinetic energy is the energy of motion • Temperature is the measure of the average kinetic energy of an object

Heat Transfer • The movement of heat from a warmer object to a colder one

Forms of heat transfer • Three forms of heat transfer: • Conduction • Convection • Radiation

Conduction • Conduction involves the transfer of heat through direct contact • Heat conductors conduct heat well, insulators do not water molecule iron atom zoomed in view

Convection • Takes place in liquids and gases as molecules move in currents • Heat rises and cold settles to the bottom

Radiation • Heat is transferred through space • Energy from the sun being transferred to the Earth

Questions • What are three types of heat transfer? • How is conduction different from radiation?

What type of heat transfer is involved? • Heating a room with a fireplace • Egg cooking in a frying pan • Roof of a house becoming hot

What type of heat transfer? • Warm air mass bringing a change in the weather • Wire getting hot from an electric appliance

Question • How is kinetic energy related to heat production?

Thermal Equilibrium Two bodies are said to be at thermal equilibrium if they are at the same temperature. If they are at different temperatures they are not in thermal equilibrium, and energy is flowing from the hot side to the cold side. hot heat cold KEhot > KEcold 26 °C KE = KE No net heat flow The two purple objects are at the same temp and, therefore are in thermal equilibrium. There is no net flow of heat energy here.

Thermometer • A instrument used to measure temperature • Thermometers commonly have alcohol (with dye) or mercury • Digital thermometers have replaced older ones • A thermometer must be allowed to come to thermal equilibrium with the object it is in contact with to determine its temperature

Temperature Scales • Fahrenheit: water freezes at 32 °F; boils at 212 °F • Celsius: water freezes at 0 °C; boils at 100 °C • Kelvin: water freezes at 273. 15 K; boils at 373. 15 K A change of 100 °C corresponds to a change of 180 °F. This means 1 C° change = 1. 8 F° change Since these scales are linear, and they’re offset by 32 °F, we get the conversion formula: F = 1. 8 C + 32

Celsius Scale • Celsius is the metric scale for measuring temperature • Based on water freezing at 0ºC and boiling at 100ºC

Kelvin scale • The Kelvin scale is a metric temperature scale measured in Kelvin units (K) where 0 K is absolute zero (or no kinetic energy) • Formula (273+ºC)= Kelvin One step on the Kelvin scale is the same as one step on the Celsius scale. These scales are off by 273 K, so: K = C + 273 Room temperature is around 293 kelvins, which is 20 °C, or 68 °F.

Absolute zero • The temperature in which all molecular motion stops (0 K)

Questions • What is the formula for converting a Celsius temperature to a Kelvin temperature? • What is the boiling point of water on the Kelvin scale? • What is the freezing point of water on the Kelvin scale?

Questions • Describe absolute zero. • What is absolute zero on the Celsius scale? -273 0 C

Practice Problems • Practice Problems 1 -3 pg 247 • Due Monday 3/31

12. 1 Heat and Thermal Energy Objectives • Distinguish heat from thermal energy • Define Specific Heat • Calculate temperature changes due to heat transfer

Measuring Heat • Addition of heat • Causes an increase in temperature • Removal of heat • Causes a decrease in temperature

Calories • Unit for measuring heat • The amount of heat needed to raise 1 gram of water one degree Celsius

Temperature Changes • Joule is another unit for measuring amount heat (heat is just energy) • Mass and type of substance determine the amount of temperature change

Units for Measuring Heat Flow 2) u The calorie is also related to the Joule, the SI unit of heat and energy • named after James Prescott Joule • 4. 184 J = 1 cal Heat Capacity - the amount of heat needed to increase the temperature of an object exactly 1 o. C • Depends on both the object’s mass and its chemical composition 29

Specific Heat • The ability of a substance to absorb heat energy (specific heat) • Different substances absorb heat at different rates • The greater the mass of the object the more heat is absorbed

Heat Capacity and Specific Heat u. Specific Heat Capacity (abbreviated “C”) - the amount of heat it takes to raise the temperature of 1 gram of the substance by 1 o. C • often called simply “Specific Heat” has a HUGE value, when it is compared to other substances u. Water 31

Heat Capacity and Specific Heat u. For water, C = 4. 18 J/(g o. C) in Joules, and C = 1. 00 cal/(g o. C) in calories. u. Thus, for water: • it takes a long time to heat up, and • it takes a long time to cool off! u. Water is used as a coolant and regulates Earths temperatures! 32

Heat Capacity and Specific Heat u. To calculate, use the formula: Q = mass (in grams) x T x C uheat is abbreviated as “q” u T = change in temperature u. C = Specific Heat • Units are either: J/(g o. C) or cal/(g o. C) 33

Specific Heat Equation Q = m C T Q = thermal energy m = mass C = specific heat T = change in temp Ex: The specific heat of silicon is 703 J / (kg · ºC). How much energy is needed to raise a 7 kg chunk of silicon 10 ºC ? answer: Q = 7 kg · 703 J · 10 ºC = 49 210 J kg · ºC Note that the units do indeed work out to be energy units. *Use table 12 -1 on p. 248 for some specific heats of different substances.

Questions • How can heat be measured? • What is the unit used to measure heat? • What is specific heat?

Homework • Practice Problems 5 -8 page 248 • Due Wed. 4/2/14 *Remember T is the same in o. C or K (eg: a change of 10 o. C is a change of 10 K) So you can use specific heats in J/kg-K or J/Kg-C they are the same number

12. 1 Calorimetry: Measuring Specific Heat Objectives • Use the law of conservation of energy to determine temperature changes using calorimetry • Use Calorimetry to determine specific heats

What is a calorimeter? • Device used to measure changes in thermal energy • It can be used to measure heat given off during chemical reactions (measure energy required to burn off a food product)

Calorimetry A horseshoe at 275 ºC, is dropped into bucket of water and covered. The bucket and cover are made of an insulating material. The bucket contains 2. 5 L of water originally at 25 ºC. The 1. 9 kg shoe is made of iron, which has a specific heat of 448 J / (kg · ºC). Let’s find the temp of the horseshoe and water once equilibrium is reached. At thermal equilibrium the water and shoe are at the same temp. The total thermal energy in the bucket does not change, but it is redistributed. continued on next slide

Calorimetry (cont. ) Let T = the equilibrium temperature. Q lost by iron = Q gained by water or Eiron + Ewater = 0 miron Ciron Tiron + mwater Cwater Twater = 0 miron Ciron (Tf-Ti, iron)+ mwater Cwater (Tf-Ti, water) = 0 We’ve got a simple linear equation in which we want to solve for Tf. Isolating Tf to one side gives: miron Ciron Ti, iron + mwater Cwater Ti, water Tf = ______________ miron Ciron+ mwater Cwater *You can use this for determining Tf of any substance

Calorimetry Tf = (cont. ) miron Ciron Ti, iron + mwater Cwater Ti, water ______________ miron Ciron+ mwater Cwater (1. 9 kg) (448 J / kg · ºC) (275 ºC) + (2. 5 kg) (4186 J / kg · ºC) (25 ºC) Tf = ___________________________ (1. 9 kg) (448 J / kg · ºC) + (2. 5 kg) (4186 J / kg · ºC) Tf = 43. 8 0 C We’ve got a simple linear equation in which we want to solve for Tf. Solving it gives us Tf = 43. 8 ºC. This is the equilibrium temp--the final temp for both the shoe and water.

Practice Problems • Practice Problems 9 -12, page 252 • Due Thurs 4/3

12. 2 Changes of State Objectives • Define heats of fusion and vaporization • Calculate heat transfers needed to cause changes of state

States of Matter • Solid, Liquid and Gas (won’t discuss plasma’s yet……) • States change when the bonds between particles (atoms/molecules) are over come by thermal energy of the particles

Heat and Changes of State (phase change) • A phase change is a physical change that requires a change in heat energy • Addition or removal of HEAT

Latent Heat The word “latent” comes from a Latin word that means “to lie hidden. ” When a substance changes phases (liquid solid or gas liquid) energy is transferred without a change in temperature. This “hidden energy” is called latent heat. For example energy is required to change 0 ºC ice into 0 ºC water. When frozen, water molecules are in a crystalline structure, and energy is needed to break this structure. The energy needed is called the latent heat of fusion. Additional energy is also needed to change water at 100 ºC to steam at 100 ºC, and this is called the latent heat of vaporization.

Heat in Changes of State 1. Heat of Fusion (Hf. ) = the energy required for one kilogram of a substance to melt from a solid to a liquid Q = m. Hf. (no temperature change) Values given in Table 12 -2, page 254 2. Heat of Solidification ( - Hf. ) = the heat lost when one kilogram of liquid solidifies (or freezes) to a solid Q = - m. Hf. (no temperature change) 47

Heat in Changes of State • Heat absorbed by a melting solid is equal to heat lost when a liquid solidifies – Thus, Hfus. = - Hsolid. • Note Table 12 -2, page 254 for the number values. There is no value listed for the heat of solidification it is just -Hf 48

Heats of Vaporization and Condensation • When liquids absorb heat at their boiling points, they become vapors. 3. Heat of Vaporization (Hv. ) = the amount of heat necessary to vaporize one kilogram of a given liquid. Q = m. Hv. (no temperature change) • Table 12 -2, page 254 49

Heats of Vaporization and Condensation • Condensation is the opposite of vaporization. 4. Heat of Condensation (Hcond. ) = amount of heat released when one kilogram of vapor condenses to a liquid Q = m. Hcond. (no temperature change) • Hcond = - Hv • . or …. . Q = -m. Hv 50

Summary: Latent Heat Formulas Q = m Hf or Q = m Hv Q = thermal energy m = mass H = heat of fusion or vaporization H is the energy per unit mass needed to change the state of a substance from solid to liquid or from liquid to gas. Ex: Hf (the latent heat of fusion) for gold is 6440 J / kg. Gold melts at 1063 ºC. What is the heat energy needed to melt 5 grams of solid gold at 1063 0 C? (6440 J / kg) (0. 005 kg) = 32 J. The liquid gold will still be at 1063 ºC.

Latent Heat / Specific Heat Example Superman vaporizes a 1800 kg ice monster with his heat ray vision. The ice monster was at -20 ºC. After being vaporized he is steam at 135 ºC. How much energy did Superman expend? Substance Specific Heat (in J / kg · ºC) ice 2090 liquid water 4186 steam 1970 For water : Hf = 3. 33 · 105 J / kg; Hv = 2. 26 · 106 J / kg Q = (1800 kg) (2090 J / kg · ºC) (20 ºC) heating ice to melting pt. + (1800 kg) (3. 33 · 105 J / kg) ice to water, const. temp of 0 ºC +(1800 kg) (4186 J / kg · ºC) (100 ºC) heating water to boiling pt. + (1800 kg) (2. 26 · 106 J / kg) water to steam, const. temp of 100 ºC + (1800 kg) (1970 J / kg · ºC) (35 ºC) heating steam to 135 ºC = 5. 62 · 109 J total energy expended by Superman

Practice Problems • Practice Problems 13 -15 • Due Monday 4/7

Laws of Thermodynamics (examples upcoming) • Zeroth Law: If object A is in thermal equilibrium with object B, and if object B is in thermal equilibrium with object C, then objects A and C are also in equilibrium. This is sort of a “transitive property of heat. ” • First Law: Energy is always conserved. It can change forms: kinetic, potential, internal etc. , but the total energy is a constant. Another way to say it is that the change in thermal energy of a system is equal to the sum of the work done on it and the amount of heat energy transferred to it. • Second Law: During any natural process the total amount of entropy in the universe always increases. Entropy can be defined informally as a measure of the randomness or disorder in a system. Heat flows naturally from a hot to cooler surroundings as a consequence of the second law.

Change in Entropy Equation Because most systems are many up of so many particles, calculating entropy via probabilities would be very difficult. Fortunately, we are normally concerned only with changes in entropy. If we have a system in which energy is not changing forms, the change in entropy is defined as: Q S = T S = change in entropy Q = change in internal energy (heat flow) T = absolute temperature The 2 nd Law of Thermodynamics says that during any process: Suniverse = Ssystem + Ssurroundings 0

Change in Entropy Example A glass rod is heated and then blown by a glassblower. When it is at 185°C it is brought outside to cool. 3200 J of heat are transferred from the glass to the air, which is at 18°C. Find the change in entropy of the universe: Suniverse = Ssystem + Ssurroundings = Sglass + Sair Qglass Qair + = T Tair glass -3200 J = + 458 K 291 K = -7 J / K + 11 J / K = + 4 J / K

Entropy & Fluids Suppose a beaker of very hot water is poured into an aquarium of cool water. Conservation of energy would not be violated if all the hot water remained right at the spot where it was poured. But the 2 nd Law demands that thermal energy eventually become evenly distributed. The cool water has molecules moving at a wide range of speeds (red = fast; blue = slow). Since the water is cool, there are more blues than reds. The hot water poured in has mostly red. The aquarium has less disorder (entropy) when all the fast molecules are in one spot than when they are mixed in. With time a much more likely situation exists, with a much higher entropy. continued time

Heat Engines A heat engine takes advantage of temp differences to produce useful work. The amount of work done depends on the size of the reservoirs, engine efficiency, and the temp difference (TH - TC). QH is the heat that flows from the hot region; QC is the heat flowing into the cold region. W is the useful work done by engine. The smaller QC is, the more efficient the engine is. The engine on the right satisfies the 1 st Law but violates the 2 nd Law, i. e. , 100% efficiency is unattainable. Hot Reservoir, TH QH W Hot Reservoir, TH QH Engine QC QC = 0 Cold Reservoir, TC Real engine. QH = QC + W W Cold Reservoir, TC Impossible engine. QH = W

Refrigerators A refrigerator forces heat from a cold region into a warmer one. It takes work to do this, otherwise the 2 nd Law would be violated. Can a fridge be left open in the summer to provide a make shift air conditioner? Nope, since all heat pumped out of the fridge is pumped back into the kitchen. Since QH > QC because of the work done, leaving the refrigerator open would actually make your house hotter! Hot Reservoir, TH QH W Hot Reservoir, TH QH Engine QC QC Cold Reservoir, TC Real fridge. QC + W = QH W=0 Cold Reservoir, TC Impossible fridge. QC = QH