# Ch. 6 – Thermal Energy I. Temperature and Energy Transfer Temperature

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Ch. 6 – Thermal Energy I. Temperature and Energy Transfer Temperature
Heat Transfer

A. Temperature Temperature
measure of the average KE of the particles in a sample of matter Thermometers rely on expansion

B. Temperature Scales Fahrenheit and Celsius most familiar to you
Celsius used often in science – why? Kelvin, based on Celsius and absolute zero Absolute zero = temperature at which an object’s energy is the lowest Celsius – Kelvin Conversion K = oC + 273 ? K = 32oC K = 32oC = 305K

Answers to Homework p. 328 1. a. -423.2 oF 20K b. -40.0 oF 233K
c oF 1337K 2. See table 3. D 4. C Ex oC oF K LR Air 21 70 294 Metal in car 115 239 388 Liquid N -200 -328 73 Desert air 43 110 316

C. Energy Transfer Molecules of warmer substance are moving faster than molecules of colder substance As these substances come in contact with each other, energy is transferred from warm to cold objects because of collisions of particles Example: ice cube in a warm hand

D. Thermal Energy Thermal Energy
the total energy of the particles in a material KE - movement of particles PE - forces within or between particles due to position depends on temperature, mass, and type of substance

D. Thermal Energy A B Which beaker of water has more thermal energy?
B - same temperature, more mass 200 mL 80ºC A 400 mL B

E. Heat Transfer Heat Like work, heat is...
thermal energy that flows from a warmer material to a cooler material Like work, heat is... measured in joules (J) a transfer of energy

A B E. Heat Transfer A B Why does A feel hot and B feel cold? 80ºC
Heat flows from A to your hand = hot Heat flows from your hand to B = cold 80ºC A 10ºC B A B

F. Thermal Energy Transfer

G. Conduction Conduction = movement of heat energy by direct contact
Good conductors = material through which energy can be easily transferred as heat Ex: metals, saltwater Opposite of conductor = insulator – slows the transfer of energy as heat Ex: cork, wood, glass, air

H. Convection Convection is the movement of gases or liquids from a warmer spot to a cooler spot A lava lamp displays convection as heated wax expands and rise and cooler wax falls

I. Radiation Electromagnetic waves - require no matter to travel through; ex: light, IR, UV Electromagnetic waves transfer energy = radiation When sunlight hits earth, its radiation is absorbed or reflected Darker surfaces absorb more of the radiation and lighter surfaces reflect the radiation

J. Thermal Energy Summary

K. Heat Transfer Specific Heat (C)
amount of energy required to raise the temp. of 1 kg of material by 1 degree Kelvin units: J/(kg·K) or J/(kg·°C) or J/(g·°C)

K. Heat Transfer 50 g Al 50 g Cu Al – It has a higher specific heat
Which sample will take longer to heat to 100°C? 50 g Al 50 g Cu Al – It has a higher specific heat Al will also take longer to cool down

q = m  C  T K. Heat Transfer q: heat (J) m: mass (kg)
C: specific heat (J/kg·K) T: change in temperature (K or °C) – q = heat loss + q = heat gain T = Tf - Ti

K. Heat Transfer heat gained = heat lost Calorimeter
Coffee cup Calorimeter Calorimeter device used to measure changes in thermal energy in an insulated system: heat gained = heat lost

K. Heat Transfer GIVEN: WORK: m = 32.0 g q = m·C·T Ti = 60.0°C
A 32.0-g silver spoon cools from 60.0°C to 20.0°C. How much heat is lost by the spoon? GIVEN: m = 32.0 g Ti = 60.0°C Tf = 20.0°C q = ? C = 235 J/kg·K WORK: q = m·C·T m = 32 g = kg T = 20°C - 60°C = – 40°C q = (0.032kg)(-40°C)(235J/kg·K) q = – 301 J

K. Heat Transfer GIVEN: WORK: m = 230. g q = m·C·T Ti = 12.0°C
How much heat is required to warm 230 g of water from 12.0°C to 90.0°C? GIVEN: m = 230. g Ti = 12.0°C Tf = 90.0°C q = ? C= 4184 J/kg·K WORK: q = m·C·T m = 230 g = 0.23 kg T = 90°C - 12°C = 78°C q = (0.23kg)(78°C)(4184 J/kg·K) q = 75,000J

K. Heat Transfer GIVEN: WORK: m = .400 kg q = m•C•T Ti = 273 K
A kg sample of glass requires 3190 J for its temperature to increase from 273 K to 308 K. What is the specific heat for this type of glass? GIVEN: m = .400 kg Ti = 273 K Tf = 308 K q = 3190 J C= ? J/kg·K WORK: q = m•C•T T = 308 K – 273 K = 35 K (3190 J) = (0.400 kg)(35 K)C C= 228 J/kg·K

K. Heat Transfer GIVEN: WORK: m = 550 g q = m·T·Cp Ti = 24°C
CCu= 0.38 J/g·K How much energy would be absorbed by 550 g of copper when it is heated from 24°C to 45°C? GIVEN: m = 550 g Ti = 24°C Tf = 45°C q = ? C= 0.38 J/g·K WORK: q = m·T·Cp T = 45°C - 24°C = 21°C q = (550g)(21°C)(0.38 J/g·K) q = 4,400 J

Assignments Homework Finish Thermal Energy WS – Work on it now so you can get some help!

Using Heat Cooling buildings, heating homes, preserving food is all because of heat transfer First Law of Thermodynamics – energy is conversed when transferred to due to work, heat or both Second Law of Thermodynamics – energy transferred always moves from high temp object to low temp object

Cooling Systems Substances that easily evaporate and condense are used in air conditioners When liquid evaporates, absorbs energy from surrounding air, cooling it Outside, gas condenses, releasing energy

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