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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 – 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 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 40• Mechanical Equivalent of Heat• Count Rumford – boring cannons• Joule (~ 1850) showed that

4.186 J = 1 calwhich is equivalent to4.186 kJ = 1 kcal

• Thus heat flow is atransfer of energy from onebody to another

• Study example 14.1

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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 40• Internal Energy of an Ideal Gas• For a system of N atoms, the internal energy, U, is:• U = N X KEave = N(½ mv2

ave) = (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 rotationalKE must be taken intoconsideration

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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 ΔT units 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 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 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