1. Ch.12 Thermal Energy
Thermal (Heat) Engine
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2. Ch.12 Thermal Energy Read Ch. 12 Sec. 12.1 p.241 – 248
Make own notes of the sections: kinetic-molecular theory of heat energy, Temperature vs Thermal energy,Celsius vs Kelvin Temperature scale
Do Practice Problems: #1- 4 p. 247
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3. Ch.12 Thermal Energy
Kelvin Scale Conversion:
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4. Ch.12 Thermal Energy Kinetic Molecular Theory:
Temperature vs. Thermal (Heat) energy:
Thermal Energy Transfer:
(Conduction, Convection, Radiation)
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5. Ch.12 Thermal Energy Conduction: Aluminum Copper Iron glass Time:
Prediction
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6. 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 are also able to lose heat just as quickly. (Metals make great conductors, but bad insulators.)
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7. Ch.12 Thermal Energy Specific heat capacity =
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.
The amount of heat energy required to heat up 1kg of that material by 1oC
Specific heat capacity =
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8. 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/kgK
Which material on your list has the smallest Cp?
Ans: lead = 130 J/kgK
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9. 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
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10. Ch.12 Thermal Energy Q = Eh = m x Cp x ΔT Where: m = mass in kg
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*oC or (J/kg*K)
ΔT = change in temperature (K or oC)
(Difference between starting and final temperature.)
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11. Ch.12 Thermal Energy Example 1:
How much heat energy is require to increase the temperature of 55 kg of iron from 25 oC to 320 oC?
Solution:
Eh = m x Cp x ΔT
= 55 x 450 x 295
= J
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12. Ch.12 Thermal Energy Example 2: ΔT = Eh/m*Cp
A 6.7 kg piece of aluminum is initially at 78 oC. Water is used to cool the aluminum and removes J of heat energy. What is the final temperature of the aluminum?
Solution:
ΔT = Eh/m*Cp
= / (6.7 x 903)
= 7.4 oC
Therefore: Tf = Ti – 7.4
Tf = 78 – 7.4 = oC
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13. Ch.12 Thermal Energy Example 3:
A 24 kg piece of copper is initially at 65 oC. A total mass of 55 kg of water initially at 10 oC 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.
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