Heat
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The document discusses various topics related to heat and temperature. It defines different temperature scales including Fahrenheit, Celsius, and Kelvin. It explains how to convert between temperature scales using formulas. It describes concepts such as thermal capacity, specific heat capacity, phases of matter, phase changes, and latent heats. An example problem demonstrates how to calculate the heat required to change the phase of water from liquid to vapor.
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Heat
- 2. Temperature scales degree Fahrenheit ยฐF 1724 degree Celcius ยฐC 1742 Kelvin K 1848 212 100 37.8 0 -17.78 32 Heat 2 -459.67 -273.15 0 100 373.15 310.93 255.37 0 273.15 ํํํถํ โ 2ํป2ํ Absolute zero
- 3. Temperature conversion ํ ยฐํถ = ํ ํพ โ 273.15 ํ ํพ = ํ ยฐํถ + 273.15 Heat 3 ํ ยฐํถ = 5 9 โ ํ ยฐํน โ 32 ํ ยฐํน = 9 5 โ ํ ยฐํถ + 32
- 4. Temperature conversion ํ ยฐํถ = ํ ํพ โ 273.15 ํ ํพ = ํ ยฐํถ + 273.15 Heat 4 ํ ยฐํถ = 5 9 โ ํ ยฐํน โ 32 ํ ยฐํน = 9 5 โ ํ ยฐํถ + 32
- 5. Cold vs. Hot: Random motion LOW ๏ฌ Kinetic energy ๏ ๏ฎ HIGH SOLID COLD HOT LIQUID Heat 5 GAS
- 6. Internal energy It takes effort (work) to increase the average distance between molecules. This is stored in the substance as potential energy. Heat 6 Translational kinetic energy Rotational kinetic energy Vibrational kinetic energy + Total kinetic energy ํน ํํํกํํํํํ ํ Expansion ํํํํกํํํํํ < 0 ฮํธํ > 0 ํ Compression ํํํํกํํํํํ > 0 ฮํธํ < 0 Total potential energy + Total internal energy
- 7. Macroscopic vs. microscopic ํฃ 2 ํฃ 7 ํ molecules ํฃ 10 ํฃ 6 ํฃ 9 ํฃ 8 ํฃ 3 ํฃ 4 ํฃ 1 ํฃ 5 ํฃ 11 Heat 7 ํ ํฃ ํ = 0 ํ ํฃ ํ 2 = ํ โ ํฃํํฃํ 2 > 0 NO macroscopic motion ํ โ 1 2 2 = ํธํํํ = measure for temperature ํํฃํํฃํ Temperature is a macroscopic quantity Temperature is a result of statistical mechanics
- 8. Transport of internal energy Insulation Heat 8 ํ ํก = 0ํ ํ1 ํ2 Loses internal energy ํ1 decreases Gains internal energy ํ2 increases ํก > 0ํ ํ1 ํ = 0 ํ2 ํ1 > ํ2 The energy transported from body 1 to body 2 is called thermal energy or heat (Q) ํ1 = ํ2 The end stage represents Thermal equilibrium Insulation
- 9. Heat & Work - I Explanation: Gravitational potential energy is converted into internal energy Heat 9 James Prescott Joule, 1847 Action: Release the crank Reaction: The water temperature increases
- 10. Heat & Work - II Heat is converted into workโฆbut never completely Heat 10
- 11. Thermal properties of matter HEAT from combustion providesโฆ Specific latent heat 3: vaporization = destruction of molecular bonds Specific heat capacity 2: increased internal energy of the water Thermal capacity 1: increased internal energy of the pot Heat 11
- 12. Thermal capacity ํถ = ํ ฮํ Heat 12 ํ( ํฝ) ํ โถํ + ฮํ The thermal capacity ํถ is the amount of (internal) energy (J) an object stores if it becomes 1ยฐC (1 K) hotter C = 200Jํพโ1 ํถ = 4 โ 108ํฝํพโ1 ํถ = 6 โ 105ํฝํพโ1 The house The air inside
- 13. Specific heat capacity Heat 13 ํ( ํฝ) ํ โถํ + ฮํ The specific heat capacity ํ is the amount of (internal) energy (J) an material per kg stores if it becomes 1ยฐC (1 K) hotter ํถ = ํ โ ํ = ํ ฮํ โ ํ = ํ ํ โ ฮํ If the body consists of one single substance/material: ํถ โ ํ or ํถ = ํ โ ํ The constant ํ is specific for this material and can be found in data tables ํํํกํํ ํ = 4.18 โ 103ํฝํพโ1ํํโ1 ํํํํขํํ ํ = ํํ2โ = ํ โ 1ํ 2 โ 0,6ํ = 1.9ํ3 ํํํ ํ ํ = ํ โ ํ = 998ํํํโ3 โ 1.9ํ3 = 1.9 โ 103ํํ ํโํํํํํ ํํํํํํํกํฆ ํํํํ ํถ = ํ โ ํ = 4.18 โ 103ํฝํพโ1ํํโ1 โ 1.9 โ 103ํํ = 8 โ 106ํฝํพโ1
- 14. Phases and phase changes GAS Vaporization Condensation LIQUID SOLID Heat 14 ํ โถ ํ โถ โถ ํ Melt / Fusion โถํ Solidification / Frost ํ Boiling point Melting point
- 15. Specific latent heat Heat required to melt/fuse 1 kg = โLatent heat of fusionโ ํฟํ , which is needed to weaken the intermolecular bonds and increase potential energy Heat required to vaporize 1 kg = โLatent heat of vaporizationโ ํฟํฃ , which is needed to break the already weakend bonds and set the molecules free Material ํํํํํก(ํพ) Lํ ํํฝํํโ1 ํํํํํ (ํพ) Lํฃ ํํฝํํโ1 Water 273.15 334 373.15 2260 Sulphur 386 39 718 1510 Nitrogen 63 26 77 199 Heat 15
- 16. Example 33.4 โ 106ํฝ Heat 16 How much energy (heat) does it take to convert 100 kg water of 10ยฐC to steam of 200ยฐC? Water Specific heat capacity (liquid) 4.18 ํํฝํํโ1ํพโ1 Specific heat capacity (vapor) 1.5 ํํฝํํโ1ํพโ1 Latent heat of fusion 334 ํํฝํํโ1 Latent heat of evaporation 2260 ํํฝํํโ1 Heat water from 20ยฐC to 100ยฐ : ํ = ํ โ ํ โ ฮํ = 4.18 โ 103 โ 100 โ 100 โ 10 = Vaporize the boiling water: ํ = ํฟํฃ โ ํ = 334 โ 103 โ 100 = Heat vapor from 100ยฐC to 200ยฐ : ํ = ํ โ ํ โ ฮํ = 1.5 โ 103 โ 100 โ 200 โ 100 = 37.62 โ 106ํฝ 15 โ 106ํฝ + 86 โ 106ํฝ
- 17. Example Heat taken up by cup & water equals heat released by the iron Heat 17 A piece of iron 200ํ 900โ is inserted in a cup C = 460 JKโ1 filled with 500mL water 18โ . At which temperature will there be thermal equilibrium? Material ํ(ํํฝํํโ1ํพโ1) Water 4.18 Iron 0.45 ํโ = ํโ ํถ โ ํํํ โ 18 + ํํคํํกํํ โ ํํคํํกํํ โ ํํํ โ 18 = ํํํํํ โ ํํํํํ โ 900 โ ํํํ 460 โ ํํํ โ 18 + 4.18 โ 103 โ 0.998 โ 0.500 โ ํํํ โ 18 = 0.45 โ 103 โ 0.200 โ 900 โ ํํํ 460 โ ํํํ โ 18 + 2086 โ ํํํ โ 18 = 90 โ 900 โ ํํํ 460 โ ํํํ โ 8280 + 2086 โ ํํํ โ 37548 = 81000 โ 90 โ ํํํ 460 โ ํํํ + 2086 โ ํํํ + 90 โ ํํํ = 81000 + 8280 + 37548 2636 โ ํํํ = 126828 ํํํ = 48โ
- 18. END Disclaimer This document is meant to be apprehended through professional teacher mediation (โlive in classโ) together with a physics text book, preferably on IB level. Heat 18