heat
DESCRIPTION
Heat & Internal energyTRANSCRIPT
Heat
Photo: Wikimedia.org
Temperature scales
Heat 2
degree Celcius
°C 1742
Kelvin
K 1848
degree Fahrenheit
°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
𝑁𝑎𝐶𝑙 ∙ 2𝐻2𝑂
Absolute zero
Temperature conversion
Heat 3
𝑇 °𝐶 = 𝑇 𝐾 − 273.15 𝑇 𝐾 = 𝑇 °𝐶 + 273.15
𝑇 °𝐶 =5
9∙ 𝑇 °𝐹 − 32 𝑇 °𝐹 =
9
5∙ 𝑇 °𝐶 + 32
Temperature conversion
Heat 4
𝑇 °𝐶 = 𝑇 𝐾 − 273.15 𝑇 𝐾 = 𝑇 °𝐶 + 273.15
𝑇 °𝐶 =5
9∙ 𝑇 °𝐹 − 32 𝑇 °𝐹 =
9
5∙ 𝑇 °𝐶 + 32
Cold vs. Hot: Random motion
Heat 5
COLD HOT
SOLID
LIQUID
GAS
LOW Kinetic energy HIGH
Internal energy
Heat 6
Translational kinetic energy
Rotational kinetic energy
Vibrational kinetic energy
+
Total kinetic energy
𝐹𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑠Expansion
𝑊𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 < 0∆𝐸𝑝 > 0
𝑠
Compression𝑊𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 > 0∆𝐸𝑝 < 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.
Macroscopic vs. microscopic
Heat 7
𝑣2 𝑣7 𝑣10 𝑣6
𝑣9
𝑣8
𝑣3 𝑣1 𝑣5
𝑣4
𝑣11
𝑁 molecules
𝑖
𝑣𝑖 = 0
𝑖
𝑣𝑖2 = 𝑁 ∙ 𝑣𝑎𝑣𝑔
2 > 0
NO macroscopic motion
𝑁 ∙ 12𝑚𝑣𝑎𝑣𝑔2 = 𝐸𝑘𝑖𝑛 =measure for temperature
Temperature is a macroscopic quantityTemperature is a result of statistical mechanics
Transport of internal energy
Heat 8
𝑇1 𝑇2𝑡 = 0𝑠𝑄
Loses internal energy𝑇1 decreases
Gains internal energy𝑇2 increases
𝑇1 𝑇2𝑡 > 0𝑠 𝑄 = 0
𝑇1 > 𝑇2
The energy transported from body 1 to body 2is called thermal energy or heat (Q)
𝑇1 = 𝑇2
The end stage representsThermal equilibrium
Insulation
Insulation
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
Heat & Work - II
Heat 10
Heat is converted into work…but never completely
Thermal properties of matter
Heat 11
HEAT from combustion provides…
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
Thermal capacity
Heat 12
𝑇 ⟶ 𝑇 + Δ𝑇𝑄( 𝐽)
The thermal capacity 𝐶 is the amount of (internal) energy (J) an object stores if it becomes 1°C (1 K) hotter
𝐶 =𝑄
Δ𝑇
C = 200J𝐾−1
𝐶 = 4 ∙ 108𝐽𝐾−1 𝐶 = 6 ∙ 105𝐽𝐾−1
The house The air inside
Specific heat capacity
Heat 13
𝑇 ⟶ 𝑇 + Δ𝑇𝑄( 𝐽)
The specific heat capacity 𝑐 is the amount of (internal) energy (J) an material per kg stores if it becomes 1°C (1 K) hotter
𝐶 = 𝑐 ∙ 𝑚 =𝑄
Δ𝑇⇒ 𝑐 =
𝑄
𝑚 ∙ ∆𝑇
If the body consists of one single substance/material:
𝐶 ∝ 𝑚 or 𝐶 = 𝑐 ∙ 𝑚The constant 𝑐 is specific for this material and can be found in data tables
𝑐 = 4.18 ∙ 103𝐽𝐾−1𝑘𝑔−1𝑊𝑎𝑡𝑒𝑟
𝑉𝑜𝑙𝑢𝑚𝑒 𝑉 = 𝜋𝑟2ℎ = 𝜋 ∙ 1𝑚 2 ∙ 0,6𝑚 = 1.9𝑚3
𝑀𝑎𝑠𝑠 𝑚 = 𝜌 ∙ 𝑉 = 998𝑘𝑔𝑚−3 ∙ 1.9𝑚3 = 1.9 ∙ 103𝑘𝑔
𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑝𝑜𝑜𝑙 𝐶 = 𝑐 ∙ 𝑚 = 4.18 ∙ 103𝐽𝐾−1𝑘𝑔−1 ∙ 1.9 ∙ 103𝑘𝑔 = 8 ∙ 106𝐽𝐾−1
Phases and phase changes
Heat 14
SOLID
LIQUID
GAS
𝑄 ⟶
𝑄 ⟶
⟶ 𝑄
⟶ 𝑄Melt / Fusion Solidification / Frost
CondensationVaporization
𝑇
Boiling point
Melting point
Specific latent heat
Heat 15
Heat required to melt/fuse 1 kg = ‘Latent heat of fusion’ 𝐿𝑓 ,
which is needed to weaken the intermolecular bonds and increase potential energy
Heat required to vaporize 1 kg = ‘Latent heat of vaporization’ 𝐿𝑣 , which is needed to break the already weakend bonds and set the molecules free
Material 𝑇𝑚𝑒𝑙𝑡(𝐾) L𝑓 𝑘𝐽𝑘𝑔−1 𝑇𝑏𝑜𝑖𝑙(𝐾) L𝑣 𝑘𝐽𝑘𝑔
−1
Water 273.15 334 373.15 2260
Sulphur 386 39 718 1510
Nitrogen 63 26 77 199
Example
Heat 16
How much energy (heat) does it take to convert 100 kg water of 10°C to steam of 200°C?
Water
Specific heat capacity (liquid) 4.18 𝑘𝐽𝑘𝑔−1𝐾−1
Specific heat capacity (vapor) 1.5 𝑘𝐽𝑘𝑔−1𝐾−1
Latent heat of fusion 334 𝑘𝐽𝑘𝑔−1
Latent heat of evaporation 2260 𝑘𝐽𝑘𝑔−1
Heat water from 20°C to 100° : 𝑄 = 𝑐 ∙ 𝑚 ∙ ∆𝑇 = 4.18 ∙ 103 ∙ 100 ∙ 100 − 10 =
Vaporize the boiling water: 𝑄 = 𝐿𝑣 ∙ 𝑚 = 334 ∙ 103 ∙ 100 =
Heat vapor from 100°C to 200° : 𝑄 = 𝑐 ∙ 𝑚 ∙ ∆𝑇 = 1.5 ∙ 103 ∙ 100 ∙ 200 − 100 =
37.62 ∙ 106𝐽
33.4 ∙ 106𝐽
15 ∙ 106𝐽
+86 ∙ 106𝐽
Example
Heat 17
A piece of iron 200𝑔 900℃ is inserted in a cup C = 460 JK−1 filled with 500mL water 18℃ .
At which temperature will there be thermal equilibrium?
Material 𝑐(𝑘𝐽𝑘𝑔−1𝐾−1)
Water 4.18
Iron 0.45
𝑄↑ = 𝑄↓𝐶 ∙ 𝑇𝑒𝑞 − 18 + 𝑐𝑤𝑎𝑡𝑒𝑟 ∙ 𝑚𝑤𝑎𝑡𝑒𝑟 ∙ 𝑇𝑒𝑞 − 18 = 𝑐𝑖𝑟𝑜𝑛 ∙ 𝑚𝑖𝑟𝑜𝑛 ∙ 900 − 𝑇𝑒𝑞
460 ∙ 𝑇𝑒𝑞 − 18 + 4.18 ∙ 103 ∙ 0.998 ∙ 0.500 ∙ 𝑇𝑒𝑞 − 18 = 0.45 ∙ 10
3 ∙ 0.200 ∙ 900 − 𝑇𝑒𝑞
460 ∙ 𝑇𝑒𝑞 − 18 + 2086 ∙ 𝑇𝑒𝑞 − 18 = 90 ∙ 900 − 𝑇𝑒𝑞
460 ∙ 𝑇𝑒𝑞 − 8280 + 2086 ∙ 𝑇𝑒𝑞 − 37548 = 81000 − 90 ∙ 𝑇𝑒𝑞
460 ∙ 𝑇𝑒𝑞 + 2086 ∙ 𝑇𝑒𝑞 + 90 ∙ 𝑇𝑒𝑞 = 81000 + 8280 + 37548
2636 ∙ 𝑇𝑒𝑞 = 126828
𝑇𝑒𝑞 = 48℃
Heat taken up by cup & water equals heat released by the iron
END
Heat 18
DisclaimerThis document is meant to be apprehended through professional teacher mediation (‘live in class’) together with a physics text book, preferably on IB level.