Lect04 - Le cture4: Classical Illustrations of Macroscopic...

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Lecture 4, p 1 Lecture 4: Classical Illustrations of Macroscopic Thermal Effects Thermal conductivity Irreversibility Reference for Lecture 5: Elements Ch 5 References for this Lecture: Elements Ch 3,4A-C
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Lecture 4, p 2 Last time: Heat capacity Q C T Remember the 1 st Law of Thermodynamics: Q = U + W by (conservation of energy) If we add heat to a system, it can do two things: Raise the temperature (internal energy increases) Do mechanical work ( e.g. , expanding gas) How much does the temperature rise? Define heat capacity to be the amount of heat required to raise the temperature by 1 K. The heat capacity is proportional to the amount of material. It can be measured either at constant volume ( C V ) or constant pressure ( C P ). It depends on the material, and may also be a function of temperature.
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Lecture 4, p 3 Heat Capacity of a Solid The atoms in a solid behave like little balls connected by springs. Here’s one atom: It has x, y, and z springs as well as x, y, and z motion. If T is not too low, the equipartition theorem applies, and each kinetic and potential term contributes ½ kT to the internal energy: U = 3(½ kT) + 3(½ kT) = 3kT Therefore, a solid with N atoms has this heat capacity: C = 3Nk = 3nR Note: For solids (and most liquids), the volume doesn’t change much, So C P ~ C V (no work is done). The temperature dependence of C is usually much larger in solids and liquids than in gases (because the forces between atoms are more important) . x z y Equipartition often works near room temperature and above.
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Lecture 4, p 4 Heat Capacity and Specific Heat C c m = mol C c n = Q C T = Heat capacity: The heat energy required to raise the temperature of an object by 1K (=1º C). It depends on the amount of material. Units: J / K Specific heat: The heat capacity normalized to a standard amount of material (mass or moles). It only depends on the kind of material. Normalize to mass: Units: J/kg . K Normalize to moles: Units: J/mole . K “molar specific heat” Upper case “C” Lower case “c”
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Lecture 4, p 5 Demo and Exercise An m 1 = 485 gram brass block sits in boiling water ( T 1 = 100 ° C ). It is taken out of the boiling water and placed in a cup containing m 2 = 250 grams of room temperature ( T 2 = 24 ° C ) water. What is the final temperature, T F , of the system ( i.e. , when the two objects have the same T)? The specific heat of brass is c 1 = 380 J/kg . K . The specific heat of water is c 2 = 4184 J/kg . K .
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Lecture 4, p 6 Solution An m 1 = 485 gram brass block sits in boiling water ( T 1 = 100 ° C ). It is taken out of the boiling water and placed in a cup containing m 2 = 250 grams of room temperature ( T 2 = 24 ° C ) water. What is the final temperature, T F , of the system ( i.e. , when the two objects have the same T)? The specific heat of brass is
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Lect04 - Le cture4: Classical Illustrations of Macroscopic...

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