Heat

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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.