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Physics 114 – Lecture 40PowerPoint Presentation

Physics 114 – Lecture 40

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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 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 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 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 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 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

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