1 / 33

Work by Friction

Work by Friction. A box slides 10 m across a surface. A frictional force of 20 N is acting on the box. What is the work done by friction? What happened to this energy?. Work by Friction. A box slides 10 m across a surface. A frictional force of 20 N is acting on the box.

madonna-holman
Download Presentation

Work by Friction

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Work by Friction A box slides 10 m across a surface. A frictional force of 20 N is acting on the box. What is the work done by friction? What happened to this energy?

  2. Work by Friction A box slides 10 m across a surface. A frictional force of 20 N is acting on the box. What is the work done by friction? – 200 J What happened to this energy? It turned into sound and heat.

  3. Heat Energy 3U Physics

  4. Remember the 2nd Law • You can’t break even. You’re always going to get less useful energy out than was put in because some energy will be lost to entropy.

  5. The 2nd Law Entropy refers to the disorder of a system. Entropy always increases.

  6. The 2nd Law For our purposes, the 2nd Law means that heat will always move from materials at higher temperatures to materials at lower temperatures.

  7. Some definitions • Temperature is a measure of the average kinetic energy of the particles (measured in oC or K).

  8. Some definitions • Temperature is a measure of the average kinetic energy of the particles (measured in oC or K). • Thermal energy is the total kinetic energy and potential energy of the particles of a material (measured in J).

  9. Some definitions • Temperature is a measure of the average kinetic energy of the particles (measured in oC or K). • Thermal energy is the total kinetic energy and potential energy of the particles of a material (measured in J). • Heat is a measure of the thermal energy transferred from a warm body to a cooler one.

  10. Heat Transfer Heat may be transferred by: • conduction • convection • radiation

  11. Conduction In a solid, every atom is physically bonded to its neighbours. If heat energy is supplied to one part of a solid, the atoms vibrate faster and these vibrations are passed on to the next atom, and so on:

  12. Conduction In metals, not only do the atoms vibrate more when heated, but the free electrons move around more as well. These transfer the energy much faster than just vibrations in bonds.

  13. Convection Actually, heat doesn’t rise. Heated gas or liquid rises because it is less dense than the cooler material surrounding it (which will sink to replace the rising material).

  14. Convection Note that “radiators” actually heat primarily by convection, not radiation.

  15. Radiation Radiation is the transfer of heat in the form of infrared (long wavelength) light. It travels in straight lines in every direction and can even travel through a vacuum.

  16. Radiation Warm-blooded creatures typically maintain a body temperature warmer than their surroundings and will lose heat energy by radiating infrared light.

  17. Radiation Cold-blooded creatures will not.

  18. Specific Heat Capacity The specific heat capacity (c) of a material is the amount of energy that must be added to raise the temperature of 1.0 kg of material by 1oC or 1K. c is different for different materials.

  19. Specific Heat Capacity c is different for different materials, e.g.: • csteel = 452 J/kgoC • cglass = 840 J/kgoC • cwater = 4186 J/kgoC

  20. Heat The amount of heat (Q) required to raise the temperature of a quantity m of a material by an amount DT is therefore: Q = mcDT

  21. Heat: Example 1 Calculate the energy required to raise the temperature of 5.0 kg of water by 75oC.

  22. Heat: Example 1 Calculate the energy required to raise the temperature of 5.0 kg of water by 75oC. m = 5.0 kg c = 4186 J/kgoC DT = 75oC Q = ?

  23. Heat: Example 1 Calculate the energy required to raise the temperature of 5.0 kg of water by 75oC. m = 5.0 kg Q = mcDT c = 4186 J/kgoC Q = (5.0 kg)(4186 J/kgoC)(75oC) DT = 75oC Q = 1.6 x 106 J Q = ?

  24. Conservation of Energy Heat will be transferred from a hot object to a cold object such that the heat lost by the hot object is equal to the heat gained by the cold object. Qlost = Qgained

  25. Heat Example 2 A 0.500 kg pot of hot water for tea has cooled to 40.0oC. How much boiling water must be added to raise the temperature of the tea water to 65.0oC?

  26. Heat Example 2 A 0.500 kg pot of hot water for tea has cooled to 40.0oC. How much boiling water must be added to raise the temperature of the tea water to 65.0oC? So the cold water needs its temperature raised from 40.0oC to 65oC: DTcold = 25oC

  27. Heat Example 2 A 0.500 kg pot of hot water for tea has cooled to 40.0oC. How much boiling water must be added to raise the temperature of the tea water to 65.0oC? And the hot water needs its temperature lowered from 100.0oC to 65oC: DThot = 35oC

  28. Heat Example 2 Heat lost by hot water = Heat gained by cold water mhotcDThot = mcoldcDTcold

  29. Heat Example 2 Heat lost by hot water = Heat gained by cold water mhotcDThot = mcoldcDTcold mhotDThot = mcoldDTcold

  30. Heat Example 2 Heat lost by hot water = Heat gained by cold water mhotcDThot = mcoldcDTcold mhotDThot = mcoldDTcold mhot = mcoldDTcold ÷ DThot

  31. Heat Example 2 Heat lost by hot water = Heat gained by cold water mhotcDThot = mcoldcDTcold mhotDThot = mcoldDTcold mhot = mcoldDTcold ÷ DThot mhot = (0.500 kg)(25.0oC) ÷ (35.0oC)

  32. Heat Example 2 Heat lost by hot water = Heat gained by cold water mhotcDThot = mcoldcDTcold mhotDThot = mcoldDTcold mhot = mcoldDTcold ÷ DThot mhot = (0.500 kg)(25.0oC) ÷ (35.0oC) mhot = 0.357 kg

  33. Heat: More Practice Homework: Heat Energy

More Related