Assignment 2

course Phy 202

I deleted the questions that weren't relevant to my course. Is that okay, or does it through you off?

That's actually easier for me, and if you want to do that you can. However be sure the only things you deleted are the University Physics problems, labeled 'univ' or something very much like that, and the problems that are labeled only for Principles of Physics.

Question: from Introductory Problem Set 5 # 12: Finding the conductivity given rate of energy flow, area, temperatures, thickness of wall.

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

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Your Solution:

The conductivity is the rate of conduction, divided by the area and the temperature gradient. The temperature gradient can be found by dividing the difference in temperatures by the thickness of the wall.

R= k*a* (dT/dX) (dT/dX)= change in temp/thickness

Confidence Rating:3

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Given Solution:

** 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 the proportionality equation

• rate of thermal energy conduction = conductivity * temperature gradient * area, or in symbols

• 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 between the faces of the object. The distance `dx is often denoted L. Using L instead of `dx, the preceding proportionality can be written

• R = k * `dT / L * A

We can solve this equation for the proportionality constant k to get

• k = R * L / (`dT * A).

Your Self-Critique:

ok

Your Self-Critique Rating:3

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Question: Explain in terms of proportionalities how thermal energy flow, for a given material, is affected by area (e.g., is it proportional to area, inversely proportional, etc.), thickness and temperature gradient.

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Energy flow is inversely related to area and

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Your Solution:

As area increases, energy flow is more likely to be greater, and the temperature gradient is increased. With an increase in temperature gradient comes a decrease in thickness. So energy flow is directly proportional to area and temperature, and inversely proportional to thickness.

Confidence Rating:1

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Given Solution:

** CORRECT STUDENT ANSWER WITHOUT EXPLANATION:

Energy flow is:

• directly proportional to area

• inversely propportional to thicknessand

• directly proportional to temperature gradient

Good student answer, slightly edited by instructor:

The energy flow for a given object increases if the cross-sectional area (i.e., the area perpendicular to the direction of energy flow) increases. Intuitively, this is because the more area you have the wider the path available so more stuff can move through it. By analogy a 4 lane highway will carry more cars in a given time interval than will a two lane highway. In a similar manner, energy flow is directly proportional to cross-sectional area.

Temperature gradient is the rate at which temperature changes with respect to position as we move from one side of the material to the other. That is, temperature gradient is the difference in temperature per unit of distance across the material:

• temperature gradient is `dT / `dx.

(a common error is to interpret temperature gradient just as difference in temperatures, rather than temperature difference per unit of distance).

For a given cross-sectional area, energy flow is proportional to the temperature gradient. If the difference in the two temperatures is greater then the energy will move more quickly from one side to the other.

For a given temperature difference, greater thickness `dx implies smaller temperature gradient `dT / `dx. The temperature gradient is what 'drives' the energy flow. Thus

greater thickness implies a lesser temperature gradient

the lesser temperature gradient implies less energy flow (per unit of cross-sectional area) per unit of time and we can say that

the rate of energy flow (with respect to time) is inversely proportional to the thickness.

Your Self-Critique:

I’m not sure how to elaborate on this topic the way you did. I just didn’t have much to say about it, but I think that I answered the question. Would this be enough for you?

You're thinking along the right lines. However it's best if you think through and address the things in the given solution that weren't in your solution. It's not always possible to do that in complete detail, of course, but the more you do the clearer will be your understanding. Balance that advice with time constraints; I trust your judgement.

Your Self-Critique Rating:2

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Question: 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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Your Solution:

The equation we’re working with here is the one for the linear coefficient of thermal expansion. This is the amount of expansion per unit of length, per unit of temperature. The equation we have is;

Alpha= dL/(Lo*dT)

Since the only thing we are missing is the change in length, we can manipulate the equation to say;

dL= Alpha *(Lo*dT)

dL= (.2x10^-6 C^-1)* (2.0m*5.0C)

dL= 2x10^-6 m

Therefore the expansion that this material undergoes after a temperature change of 5 C is 2x10^-6m.

Confidence Rating:3

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Given Solution:

This problem is solved using the concept of a coefficient of expansion.

The linear coefficient of thermal expansion of a material, denoted alpha, is the amount of expansion per unit of length, per unit of temperature:

• expansion per unit of length is just (change in length) / (original length), i.e.,

• expansion per unit of length = `dL / L0

Thus expansion per unit of length, per unit of temperature is (expansion per unit of length) / `dT. Denoting this quantity alpha we have

• alpha = (`dL / L0) / `dT. This is the 'explanatory form' of the coefficient of expansion. In algebraically simplified form this is

• alpha = `dL / (L0 * `dT).

In this problem we want to find the amount of the expansion. If we understand the concept of the coefficient of expansion, we understand that the amount of the expansion is the product of the coefficient of expansion, the original length and the temperature difference: If we don’t completely understand the idea, or even if we do understand it and want to confirm our understanding, we can solve the formula alpha = `dL / (L0 * `dT) for `dL and plug in our information:

• `dL = alpha * L0 * `dT = .2 * 10^-6 C^(-1) * 2.0 m * 5.0 C = 2 * 10^-6 m.

This is 2 microns, two one-thousandths of a millimeter.

By contrast the coefficient of expansion of steel is 12 * 10^-6 C^(-1); using this coefficient of expansion yields a change in length of 1.2 * 10^-4 m, or 120 microns, which is 60 times as much as for the given alloy.

Your Self-Critique:ok

Your Self-Critique Rating:3

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Question: query general phy 13.12: what is the coefficient of volume expansion for quartz, and by how much does the volume change? (Note that Principles of Physics and University Physics students do not do General Physics problems)

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Your Solution:

Our coefficient of volume expansion is 1x10^-6 C-1.

We first need to find the volume of the quartz. D= 8.75, so r= 4.38

V= 4/3 pi r^3, V= 4/3 pi (4.38)^3, V= 351.97 cm^3

The equation we need to use in one in which we can find the change in volume after the temp change.

dV= beta*Vo*dT

dV= (1x10^-6 C-1)(351.97 cm^3)(170 C)

dV= 0.0589 cm^3

Quartz changes in volume by 0.0589 cm^3 after being subjected to a temperature change of 170 C.

Confidence Rating:2

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Given Solution:

** The coefficient of volume expansion for quartz is 1 x 10^(-6) C^(-1).

The sphere has diameter 8.75 cm, so its volume is 4/3 pi r^3 = 4/3 pi ( 4.38 cm)^3 = 352 cm^3, approx..

The coefficient of volume expansion is the proportionality constant beta in the relationship `dV = beta * V0 * `dT (completely analogous to the concept of a coefficient of linear expansion).

We therefore have

`dV = beta* V0*dT = 3 x 10^(-6) C^ (-1) * 352 cm^3 * (200C - 30 C) = 0.06 cm^3 **

Your Self-Critique:

I got the right answer, but in the actual equation you worked out, you used 3x10^-6 C-1 as the coefficient of volume expansion. I used 1x10^-6 C-1, but still got the same answer. Is that a typo or did I miss something?

Your Self-Critique Rating:ok

&#This looks good. See my notes. Let me know if you have any questions. &#