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Physics 101: Lecture 26 Temperature & Thermal Expansion

Physics 101: Lecture 26 Temperature & Thermal Expansion. Today’s lecture will cover Textbook Sections 12.1 - 12.5. Internal Energy and Temperature. All objects have “internal energy” (measured in Joules) Internal energy of a system is the energy which is

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Physics 101: Lecture 26 Temperature & Thermal Expansion

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  1. Physics 101: Lecture 26Temperature & Thermal Expansion • Today’s lecture will cover Textbook Sections 12.1 - 12.5

  2. Internal Energy and Temperature • All objects have “internal energy” (measured in Joules) Internal energy of a system is the energy which is left when the total external mechanical energy (kinetic+potential) has been subtracted. Examples: kinetic energy of molecules, binding energy • Thermal internal energy is the average kinetic energy of molecules in motion. Temperature is a measure of this average kinetic energy. The amount of thermal internal energy of an objects depends on • temperature • related to average kinetic energy per molecule • how many molecules • mass • “specific heat capacity” • related to how many different ways a molecule can move • the more ways it can move, the higher the specific heat

  3. Temperature Scales Celcius Fahrenheit Kelvin 373.15 100 212 Water boils 273.15 0 32 Water freezes 0 absolute zero -273.15 -459 Conversion factors: 1 C = 9/5 F T(K) = T (C ) + 273.15

  4. Thermal Expansion Temp: T L0 Temp: T+T L • When temperature rises • molecules have more kinetic energy • they are moving faster, on the average • consequently, things tend to expand • amount of expansion depends on… • change in temperature • original length • coefficient of thermal expansion • L0 + L = L0 +  L0 T • L =  L0 T(linear expansion) • V =  L0 T (volume expansion)

  5. CORRECT T = 100 C T = 0 C M, V0 r0 = M / V0 M, V100 r100 = M / V100 < r0 Concept Question As you heat a block of aluminum from 0 C to 100 C its density 1. Increases 2. Decreases 3. Stays the same

  6. CORRECT Concept Question An aluminum plate has a circular hole cut in it. A copper ball (solid sphere) has exactly the same diameter as the hole when both are at room temperature, and hence can just barely be pushed through it. If both the plate and the ball are now heated up to a few hundred degrees Celsius, how will the ball and the hole fit (a (aluminum) > a (copper) ) ? 1. The ball won’t fit through the hole any more 2. The ball will fit more easily through the hole 3. Same as at room temperature The aluminum plate and copper ball both have different coefficients of thermal expansion. Aluminum has a higher coefficient than copper which means the aluminum plate hole will expand to be larger than the copper ball's expansion and allow more space for the ball to pass through.

  7. Object at temp T Same object at higher T: Plate and hole both get larger Imagine a plate made from 9 smaller pieces. Each piece expands. If you remove one piece, it will leave an “expanded hole” Why does the hole get bigger when the plate expands ???

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