Thermal Energy and Heat thermal energy the total kinetic and potential energy of the atoms or molecules of a substance heat the transfer of energy from.

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Presentation transcript:

Thermal Energy and Heat thermal energy the total kinetic and potential energy of the atoms or molecules of a substance heat the transfer of energy from a warm body to a colder body temperature measure of the average kinetic energy of the atoms or molecules of a substance Mr. Edmunds – Heat and Temperature

+ = 50°C KE = = PE 50°C 70°C 60°C KE < =< = PE = = 4.5 Prac. #1,2 + =

Methods of Heat Transfer Energy is transferred from a warmer body to a cooler body in three different ways: 1. conduction through a material by collision of molecules i.e. heat through a wall of a house 2. convection through a circulating path of fluid particles i.e. heat through cracks in doors or windows (air) 3. radiation energy transmitted by electromagnetic waves i.e. heat radiating off the surface of the earth 4.5 Prac. #4-9

Calculating Heat Transfer Different substances require different amounts of energy to increase the temperature of the substance. This occurs because different substances have different abilities to hold heat. The specific heat capacity is a constant which measures the amount of energy needed to raise 1 kg of a substance by 1˚C (J/kg˚C) Water has a very high capacity to hold heat. c w = 4180 J/kg˚C This means that it takes 4180 J of energy to raise 1 kg of water 1˚C. It also means that 1 kg of water releases 4180 J of energy for every 1˚C it cools. The quantity of heat gained or lost can be found using the following equation: Q = mcΔtQ – quantity of heat (J) m – mass (kg) c – specific heat capacity t – change in temperature (˚C) 4.5 Prac. #10-15

Principal of Heat Exchange When heat is transferred from one body to another the amount of heat lost by the warmer body equals the amount of heat gained by the cooler body. Q gained = Q lost Q gained + Q lost = 0 m 1 c 1 Δt 1 + m 2 c 2 Δt 2 = Prac. #16

m 1 = 0.12 kg c 1 = 780 t i = 21°C t f = ? m 2 = 0.21 kg c 1 = 4180 t i = 91°C t f = ? m 1 c 1 (t f – t i ) + m 2 c 2 (t f – t i ) = 0 (.12)(780)(t f – 21)+(.21)(4180)(t f – 91) = t f – t f – = t f = t f = 84.3 The final temperature was 84°.