Physics 114 lecture 40
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Physics 114 – Lecture 40. Chapter 14 Heat Heat Flow: Spontaneously occurs from the hotter to the colder body. Thermal equilibrium §14.1 Heat as Energy Transfer Flow of heat – 18 th century view, flow of caloric from the hotter to the colder body

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Physics 114 – Lecture 40

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Physics 114 lecture 40

Physics 114 – Lecture 40

  • Chapter 14Heat

  • Heat Flow: Spontaneously occurs from the hotter to the colder body. Thermal equilibrium

  • §14.1 Heat as Energy Transfer

  • Flow of heat – 18th century view, flow of caloric from the hotter to the colder body

  • 19th century – heat was viewed as being similar to work, which was equivalent to viewing heat as a form of energy

  • Kinetic Theory for gases: KEave = ½ mv2ave = (3/2) kT

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Physics 114 lecture 401

Physics 114 – Lecture 40

  • Definition of calorie (cal):

  • Amount of heat needed to raise 1 gram of water through a rise in temperature of 10C – more specifically from 14.5 0C to 15.5 0C

  • 1 kcal = 1000 cal is the amount of heat required to raise 1 kg of water through a temperature rise of 10 C

  • 1 kcal ≡ 1 Cal =1000 cal, the Cal being the unit of energy used by nutritionists

  • British system of units: 1 British thermal unit (Btu) is the amount of heat required to raise 1 lb of water through a rise in temperature of 1 0F

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Physics 114 lecture 402

Physics 114 – Lecture 40

  • Mechanical Equivalent of Heat

  • Count Rumford – boring cannons

  • Joule (~ 1850) showed that

    4.186 J = 1 cal

    which is equivalent to

    4.186 kJ = 1 kcal

  • Thus heat flow is a

    transfer of energy from one

    body to another

    •Study example 14.1

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Physics 114 lecture 403

Physics 114 – Lecture 40

  • §14.2 Internal Energy

  • The internal energy of a system is the sum of all the energy of each molecule or atom in that system

  • Internal energy is sometimes referred to as thermal energy

  • Temperature, Heat and Internal Energy

  • Temperature in K: a measure of KEave of the atoms and molecules in the system

  • Heat: transfer of energy from one body to another because of a temperature difference between those bodies

  • Internal Energy: sum of energy of each molecule or atom in the system

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Physics 114 lecture 404

Physics 114 – Lecture 40

  • Internal Energy of an Ideal Gas

  • For a system of N atoms, the internal energy, U, is:

  • U = N X KEave = N(½ mv2ave) = (3/2) N kT

  • With N k = n R this becomes:

  • U = (3/2) n RT

  • Note that the temperature, T, must be expressed in K

  • When this expression is used

    for molecules, the rotational

    KE must be taken into

    consideration

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Physics 114 lecture 405

Physics 114 – Lecture 40

  • §14.3 Specific Heat

  • Consider an amount of heat, Q, flowing into a body of mass, m, which produces a rise in temperature, ΔT. Experimentally it is observed that Q is directly proportional to m and ΔT, but that Q does depend on the composition of the body.

  • The specific heat of a material is defined as:

  • Q = m c ΔTunits of c cal/(g.0C)

  • Note that, from the definition of the calorie, that the above definition also defines c for water to be

    cwater = 1 cal/(g.0C)

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Physics 114 lecture 406

Physics 114 – Lecture 40

  • Study examples 14.2 and 14.3

  • §14.4 Calorimetry – Solving Problems

  • We need to exercise care in describing various types of system, which is the collection of bodies under consideration. Systems are of three main types

  • Closed System: No mass enters or leaves but heat may be exchanged with the environment

  • Open System: Mass and energy may enter or leave

  • Isolated System: Neither mass nor energy may enter or leave

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Physics 114 lecture 407

Physics 114 – Lecture 40

  • Calorimetry problems are solved by applying the principle of conservation of energy – heat energy to be specific →

  • heat lost by one body or bodies = heat gained by the other body or bodies

  • which is the same statement as

  • energy lost by one body or bodies = energyt gained by the other body or bodies

  • Heat lost by one body = m c (Ti – Tf)

  • Heat gained by one body = m c (Tf – Ti)

  • Study examples 14.4, 14.5 and 14.6

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