Query Assignment 11

course PHY 122

{nH\~assignment #011

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011. `query 1

Physics II

02-18-2007

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09:50:15

query introset 5 # 12: Finding the conductivity given rate of energy flow, area, temperatures, thickness of wall.

Describe how we find the conductivigy given the rate of energy flow, area, temperatures, and thickness of the wall?

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RESPONSE -->

Conductivity is equal to the rate of conduction divided by the area and the gradient

With the gradient being the average temperature change per unit of distance.'dT/'dx

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09:56:42

** The rate at which thermal energy is conducted across for a object made of a given substance is proportional to the temperature gradient (the rate at which temperature changes across the object), and to the cross-sectional area of the object. The conductivity is the constant of proportionality for the given substance. So we have

Rate of thermal energy condction = conductivity * temperature gradient * area, or R = k * `dT/`dx * A.

For an object of uniform cross-section `dT is the temperature difference across the object and `dx is the distance L between the faces of the object. In this case the equation is R = k * `dT / L * A and we can solve to get k = R * L / (`dT * A). **

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10:04:50

Explain how energy flow, for a given material, is affected by area (e.g., is it proportional to area, inversely proportional, etc.), thickness and temperature gradient.

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RESPONSE -->

energy flow is proportinal to the area,

The temperature gradient is inversely proprtional to the thickness, so the thickness is also inversely proprtional to the energy flow.

The temperature gradient is proportional to the energy flow.

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10:05:46

** CORRECT STUDENT ANSWER WITHOUT EXPLANATION:

Energy flow is directly proportional to area

inversely propportional to thickness

and directly proportional to temperature gradient

GOOD STUDENT ANSWER WITH EXPLANATIONS, PLUS INSTRUCTOR COMMENTARY:

The energy flow for a given material increases if the area increases. This is because the more area you have the wider a path something has to go through so more of it can move through it. Just like a 4 lane highway will carry more cars in a given time interval than a two lane highway will. So the relationship of energy flow to area is proportional.

Energy flow, however is inversely proportional to thickness. This is because although the thermal energy flows through the material, the material impedes it. So if the thickness increases the thermal energy will have to travel farther through the resistance and be impeded more.

** Also for given temperature difference, greater thickness `dx implies smaller temperature gradient `dT / `dx. Small temperature gradient doesn't 'drive' the energy flow as much.

Energy flow is also proportional to the temperature gradient. Meaning if the difference in the two temperatures is greater then the energy will move faster from one side to the other.

Temperature gradient is not difference in temperatures, it's difference in temperature per unit of distance across the material. Temperature gradient is `dT / `dx, not just `dT.

Greater temperature gradient means greater difference in temperature over any given small distance increment. The greater the temperature difference across this increment the more energy will flow. **

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RESPONSE -->

ok

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10:18:34

principles of physics and general college physics 13.8: coeff of expansion .2 * 10^-6 C^-1, length 2.0 m. What is expansion along length if temp increases by 5.0 C?

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RESPONSE -->

'dL=(coeffecient of expansion)*L*'dT

(.2*10^-6C^-1)*2m*5C= 2*10^-6m

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10:19:04

The amount of the expansion is the product of the coefficient of expansion, the original length and the temperature difference:

`dL = alpha * L0 * `dT = .2 * 10^-6 C^(-1) * 2.0 m * 5.0 C = 2 * 10^-6 m. This is 2 microns, two-onethousandths of a millimeter.

By contrast the coefficient of expansion of steel is 12 * 10^-6 C^(-1);using this for the coefficient of expansion yields a change in length of 24 * 10^-6 m, or 24 microns, which is 240 times as much as for the given alloy.

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RESPONSE -->

ok

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Good work. Let me know if you have questions.