Ch 12 Thermal Energy Thermal Heat Engine 12162021

Ch. 12 Thermal Energy Thermal (Heat) Engine 12/16/2021 1

Ch. 12 Thermal Energy n n Read Ch. 12 Sec. 12. 1 p. 241 – 248 Make own notes of the sections: kineticmolecular theory of heat energy, Temperature vs Thermal energy, Celsius vs Kelvin Temperature scale n Do Practice Problems: #1 - 4 p. 247 12/16/2021 2

Kelvin Scale Conversion: 12/16/2021 Ch. 12 Thermal Energy 3

Ch. 12 Thermal Energy Kinetic Molecular Theory: Temperature vs. Thermal (Heat) energy: Thermal Energy Transfer: (Conduction, Convection, Radiation) 12/16/2021 4

Ch. 12 Thermal Energy n Conduction: Aluminum Copper Iron glass Time: Prediction 12/16/2021 5

Ch. 12 Thermal Energy Specific Heat Capacity (Cp) Different materials are able to absorb (and release) heat energy at different rates. Most metals are good conductors of heat and are able to heat up quickly and therefore also able to lose heat just as quickly. (Metals make great conductors, but bad insulators. ) 12/16/2021 6

Ch. 12 Thermal Energy Some materials require more energy to heat up and take longer to cool. In general the longer it takes for a material to heat up or cool down the larger the specific heat capacity for that material. Specific heat capacity 12/16/2021 = The amount of heat energy required to heat up 1 kg of that material by 1 o. C 7

Ch. 12 Thermal Energy Look on p. 248 of your text book Which material on your list has the largest Cp? Ans: Water = 4180 J/kg. K Which material on your list has the smallest Cp? Ans: lead = 130 J/kg. K 12/16/2021 8

Ch. 12 Thermal Energy Which material would require the most energy to heat up? Ans: Water = 4180 J/kg*K Which material would retain the most energy upon heating? Ans: Water = 4180 J/kg*K Which material would be the worst insulator? Ans: Lead = 130 J/kg*K 12/16/2021 9

Ch. 12 Thermal Energy This is the heat energy equation: Q = Eh = m x Cp x ΔT Where: m = mass in kg Cp = specific heat capacity in J/kg*o. C or (J/kg*K) ΔT = change in temperature (K or o. C) (Difference between starting and final temperature. ) 12/16/2021 10

Ch. 12 Thermal Energy Example 1: How much heat energy is require to increase the temperature of 55 kg of iron from 25 o. C to 320 o. C? Solution: Eh = m x Cp x ΔT = 55 x 450 x 295 = 7300000 J 12/16/2021 11

Example 2: Ch. 12 Thermal Energy A 6. 7 kg piece of aluminum is initially at 78 o. C. Water is used to cool the aluminum and removes 45000 J of heat energy. What is the final temperature of the aluminum? Solution: ΔT = Eh/m*Cp = 45000 / (6. 7 x 903) = 7. 4 o. C Therefore: Tf = Ti – 7. 4 Tf = 78 – 7. 4 = 70. 6 o. C 12/16/2021 12

Example 3: Ch. 12 Thermal Energy A 24 kg piece of copper is initially at 65 o. C. A total mass of 55 kg of water initially at 10 o. C is used to cool the copper. What is the final temperature of the copper and water? (Solution will be shown on overhead. ) Thermal or Heat Energy Go on to the worksheet. Get ready for Lab Test next class. 12/16/2021 13