Download - Heat
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Heat
Photo: Wikimedia.org
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Temperature scales
Heat 2
degree Celcius
Β°C 1742
Kelvin
K 1848
degree Fahrenheit
Β°F 1724
-273.15
0
-459.67
100
373.15
212
310.93
37.8
100
255.37
-17.78
0
0
32
273.15
πππΆπ β 2π»2π
Absolute zero
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Temperature conversion
Heat 3
π Β°πΆ = π πΎ β 273.15 π πΎ = π Β°πΆ + 273.15
π Β°πΆ =5
9β π Β°πΉ β 32 π Β°πΉ =
9
5β π Β°πΆ + 32
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Temperature conversion
Heat 4
π Β°πΆ = π πΎ β 273.15 π πΎ = π Β°πΆ + 273.15
π Β°πΆ =5
9β π Β°πΉ β 32 π Β°πΉ =
9
5β π Β°πΆ + 32
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Cold vs. Hot: Random motion
Heat 5
COLD HOT
SOLID
LIQUID
GAS
LOW Kinetic energy HIGH
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Internal energy
Heat 6
Translational kinetic energy
Rotational kinetic energy
Vibrational kinetic energy
+
Total kinetic energy
πΉπππ‘πππππ π Expansion
ππππ‘πππππ < 0βπΈπ > 0
π
Compressionππππ‘πππππ > 0βπΈπ < 0
Total potential energy
+
Total internal energy
It takes effort (work) to increase the average distance between molecules.This is stored in the substance as potential energy.
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Macroscopic vs. microscopic
Heat 7
π£2 π£7 π£10 π£6
π£9
π£8
π£3 π£1 π£5
π£4
π£11
π molecules
π
π£π = 0
π
π£π2 = π β π£ππ£π
2 > 0
NO macroscopic motion
π β 12ππ£ππ£π2 = πΈπππ =measure for temperature
Temperature is a macroscopic quantityTemperature is a result of statistical mechanics
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Transport of internal energy
Heat 8
π1 π2π‘ = 0π π
Loses internal energyπ1 decreases
Gains internal energyπ2 increases
π1 π2π‘ > 0π π = 0
π1 > π2
The energy transported from body 1 to body 2is called thermal energy or heat (Q)
π1 = π2
The end stage representsThermal equilibrium
Insulation
Insulation
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Heat & Work - I
Heat 9
James Prescott Joule, 1847
Action: Release the crank
Reaction: The water temperature increases
Explanation: Gravitational potential energy is converted into internal energy
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Heat & Work - II
Heat 10
Heat is converted into workβ¦but never completely
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Thermal properties of matter
Heat 11
HEAT from combustion providesβ¦
1: increased internal energy of the pot
2: increased internal energy of the water
3: vaporization = destruction of molecular bonds
Thermal capacity
Specific heat capacity
Specific latent heat
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Thermal capacity
Heat 12
π βΆ π + Ξππ( π½)
The thermal capacity πΆ is the amount of (internal) energy (J) an object stores if it becomes 1Β°C (1 K) hotter
πΆ =π
Ξπ
C = 200JπΎβ1
πΆ = 4 β 108π½πΎβ1 πΆ = 6 β 105π½πΎβ1
The house The air inside
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Specific heat capacity
Heat 13
π βΆ π + Ξππ( π½)
The specific heat capacity π is the amount of (internal) energy (J) an material per kg stores if it becomes 1Β°C (1 K) hotter
πΆ = π β π =π
Ξπβ π =
π
π β βπ
If the body consists of one single substance/material:
πΆ β π or πΆ = π β πThe constant π is specific for this material and can be found in data tables
π = 4.18 β 103π½πΎβ1ππβ1πππ‘ππ
ππππ’ππ π = ππ2β = π β 1π 2 β 0,6π = 1.9π3
πππ π π = π β π = 998πππβ3 β 1.9π3 = 1.9 β 103ππ
πβπππππ πππππππ‘π¦ ππππ πΆ = π β π = 4.18 β 103π½πΎβ1ππβ1 β 1.9 β 103ππ = 8 β 106π½πΎβ1
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Phases and phase changes
Heat 14
SOLID
LIQUID
GAS
π βΆ
π βΆ
βΆ π
βΆ πMelt / Fusion Solidification / Frost
CondensationVaporization
π
Boiling point
Melting point
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Specific latent heat
Heat 15
Heat required to melt/fuse 1 kg = βLatent heat of fusionβ πΏπ ,
which is needed to weaken the intermolecular bonds and increase potential energy
Heat required to vaporize 1 kg = βLatent heat of vaporizationβ πΏπ£ , which is needed to break the already weakend bonds and set the molecules free
Material πππππ‘(πΎ) Lπ ππ½ππβ1 πππππ(πΎ) Lπ£ ππ½ππ
β1
Water 273.15 334 373.15 2260
Sulphur 386 39 718 1510
Nitrogen 63 26 77 199
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Example
Heat 16
How much energy (heat) does it take to convert 100 kg water of 10Β°C to steam of 200Β°C?
Water
Specific heat capacity (liquid) 4.18 ππ½ππβ1πΎβ1
Specific heat capacity (vapor) 1.5 ππ½ππβ1πΎβ1
Latent heat of fusion 334 ππ½ππβ1
Latent heat of evaporation 2260 ππ½ππβ1
Heat water from 20Β°C to 100Β° : π = π β π β βπ = 4.18 β 103 β 100 β 100 β 10 =
Vaporize the boiling water: π = πΏπ£ β π = 334 β 103 β 100 =
Heat vapor from 100Β°C to 200Β° : π = π β π β βπ = 1.5 β 103 β 100 β 200 β 100 =
37.62 β 106π½
33.4 β 106π½
15 β 106π½
+86 β 106π½
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Example
Heat 17
A piece of iron 200π 900β is inserted in a cup C = 460 JKβ1 filled with 500mL water 18β .
At which temperature will there be thermal equilibrium?
Material π(ππ½ππβ1πΎβ1)
Water 4.18
Iron 0.45
πβ = πβπΆ β πππ β 18 + ππ€ππ‘ππ β ππ€ππ‘ππ β πππ β 18 = πππππ β πππππ β 900 β πππ
460 β πππ β 18 + 4.18 β 103 β 0.998 β 0.500 β πππ β 18 = 0.45 β 10
3 β 0.200 β 900 β πππ
460 β πππ β 18 + 2086 β πππ β 18 = 90 β 900 β πππ
460 β πππ β 8280 + 2086 β πππ β 37548 = 81000 β 90 β πππ
460 β πππ + 2086 β πππ + 90 β πππ = 81000 + 8280 + 37548
2636 β πππ = 126828
πππ = 48β
Heat taken up by cup & water equals heat released by the iron
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END
Heat 18
DisclaimerThis document is meant to be apprehended through professional teacher mediation (βlive in classβ) together with a physics text book, preferably on IB level.