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. What is the work done by friction? 200 J What happened to this energy? It turned into sound and heat.
Heat Energy 3U Physics
Remember the 2 nd 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.
The 2 nd Law Entropy refers to the disorder of a system. Entropy always increases.
The 2 nd Law For our purposes, the 2 nd Law means that heat will always move from materials at higher temperatures to materials at lower temperatures.
Some definitions Temperature is a measure of the average kinetic energy of the particles (measured in o C or K).
Some definitions Temperature is a measure of the average kinetic energy of the particles (measured in o C or K). Thermal energy is the total kinetic energy and potential energy of the particles of a material (measured in J).
Some definitions Temperature is a measure of the average kinetic energy of the particles (measured in o C 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.
Heat Transfer Heat may be transferred by: conduction convection radiation
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:
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. Heating a Solid Animation
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).
Convection Note that radiators actually heat primarily by convection, not radiation.
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.
Radiation Warm-blooded creatures typically maintain a body temperature warmer than their surroundings and will lose heat energy by radiating infrared light.
Cold-blooded creatures will not. Radiation
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 1 o C or 1K. c is different for different materials.
Specific Heat Capacity c is different for different materials, e.g.: c steel = 452 J/kg o C c glass = 840 J/kg o C c water = 4186 J/kg o C
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
Heat: Example 1 Calculate the energy required to raise the temperature of 5.0 kg of water by 75 o C.
Heat: Example 1 Calculate the energy required to raise the temperature of 5.0 kg of water by 75 o C. m = 5.0 kg c = 4186 J/kg o C DT = 75 o C Q =?
Heat: Example 1 Calculate the energy required to raise the temperature of 5.0 kg of water by 75 o C. m = 5.0 kg Q = mcdt c = 4186 J/kg o C Q = (5.0 kg)(4186 J/kg o C)(75 o C) DT = 75 o C Q = 1.6 x 10 6 J Q =?
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. Q lost = Q gained
Heat Example 2 A 0.500 kg pot of hot water for tea has cooled to 40.0 o C. How much boiling water must be added to raise the temperature of the tea water to 65.0 o C?
Heat Example 2 A 0.500 kg pot of hot water for tea has cooled to 40.0 o C. How much boiling water must be added to raise the temperature of the tea water to 65.0 o C? So the cold water needs its temperature raised from 40.0 o C to 65 o C: DT cold = 25 o C
Heat Example 2 A 0.500 kg pot of hot water for tea has cooled to 40.0 o C. How much boiling water must be added to raise the temperature of the tea water to 65.0 o C? And the hot water needs its temperature lowered from 100.0 o C to 65 o C: DT hot = 35 o C
Heat Example 2 Heat lost by hot water = Heat gained by cold water m hot cdt hot = m cold cdt cold
Heat Example 2 Heat lost by hot water = Heat gained by cold water m hot cdt hot = m cold cdt cold m hot DT hot = m cold DT cold
Heat Example 2 Heat lost by hot water = Heat gained by cold water m hot cdt hot = m cold cdt cold m hot DT hot = m cold DT cold m hot = m cold DT cold DT hot
Heat Example 2 Heat lost by hot water = Heat gained by cold water m hot cdt hot = m cold cdt cold m hot DT hot = m cold DT cold m hot = m cold DT cold DT hot m hot = (0.500 kg)(25.0 o C) (35.0 o C)
Heat Example 2 Heat lost by hot water = Heat gained by cold water m hot cdt hot = m cold cdt cold m hot DT hot = m cold DT cold m hot = m cold DT cold DT hot m hot = (0.500 kg)(25.0 o C) (35.0 o C) m hot = 0.357 kg
Heat: More Practice Homework: Heat Energy