Resistance temperature coefficient

[cycrups]

Homework Statement

1. In the physics lab the following values of a resistor have been recorded as a function of the temperature:

Temperature T/°C
20
30
40
50
60

Resistance R/

213
143
98
69
47It is said that the temperature dependence of R can be described by the following equation: R(1/T)=a*exp(b/T) , with T as absolute temperature in Kelvin.

1. Plot the natural logarithm of R as a function of 1/T and find from that graph the parameters a, and b. Use the graph paper at the last page!
2. Do you agree that the above equation gives a good description of the data?
3. Which value of R do you expect for a temperature of 100°C?

Homework Equations

R(1/T)=a*exp(b/T)

The Attempt at a Solution

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So I drew the graph on excel http://imgur.com/RPowO0o (in Kelving of course because that's what he asked for)
What I don't know is how will I find a and b if there are two unknowns in one formula. I have though of b and a to be the mean of each the Temp and the Resistance in the graph as the teacher stated I should be able to solve them from the graph but I have no idea.

 

[Chestermiller]

You didn't plot the data the way your professor told you to plot it (in item # 1). Do that first, and see what you get.

Chet

 

[BvU]

You also want to check why the data points on your R(T) plot aren't evenly spaced in T

 

[cycrups]

You didn't plot the data the way your professor told you to plot it (in item # 1). Do that first, and see what you get.

Chet

How can I plot a natural log of R as a function of 1/T. I don't get it. Can you please explain how I do it?

 

[Chestermiller]

How can I plot a natural log of R as a function of 1/T. I don't get it. Can you please explain how I do it?

You have a column of numbers with the values of T, and, right next to it, you have a column of numbers with the corresponding values of R. Now, in a third column, right next to these two, fill in the numbers 1/T, where T is the absolute temperature. Next, in a fourth column next to these three, fill in the numbers lnR corresponding to the R values. You are going to make a graph of the third and forth columns on the piece of graph paper they provided. Does this make sense?

Chet

 

[cycrups]

You have a column of numbers with the values of T, and, right next to it, you have a column of numbers with the corresponding values of R. Now, in a third column, right next to these two, fill in the numbers 1/T, where T is the absolute temperature. Next, in a fourth column next to these three, fill in the numbers lnR corresponding to the R values. You are going to make a graph of the third and forth columns on the piece of graph paper they provided. Does this make sense?

Chet

Yes it does. I redrew the graph with ln R and 1/T http://i.imgur.com/q5s9oLJ.png is this correct and what's the next step to figure out a and b?

 

[BvU]

Veritcally I see ln R plotted. Good. But horizontally, I still see T ...

 

[henxan]

Yes it does. I redrew the graph with ln R and 1/T http://i.imgur.com/q5s9oLJ.png is this correct and what's the next step to figure out a and b?

After you get your x-axis straightened out you can imagine the that the function $a\cdot e^{\frac{b}{T}}$ is a model that describes how your graph works. It can be extended in both directions. If you go to the right $\frac{1}{T}$ increases - which in turn corresponds to a decrease in T, and vice versa. Remember, your y-axis is now the ln(R), where negative values mean R less than 1 and positive values mean R higher than 1 - R will never go negative ;)..

You will observe that if you extend a line from your graph to the left, corresponding to a temperature increase, it will eventually intersect with x- and y-axis. Imagine that your temperature increases to infinity $T \rightarrow \infty$, this means that $\frac{1}{T}\rightarrow 0$ (You will of course never reach this point). What value will $a\cdot e^{\frac{b}{T}}$ have then? This will solve one of your variables.

Further if your line crosses the x-axis which is situated at y=0 (which corresponds to R=1), what value must your other variable have then? ;)
Hint: Solve $R = a\cdot e^{\frac{b}{T}}$ where you now only have 1 unknown variable.

 

[Chestermiller]

Yes it does. I redrew the graph with ln R and 1/T http://i.imgur.com/q5s9oLJ.png is this correct and what's the next step to figure out a and b?

I don't see 1/T on your graph. So this is not correct. Make the horizontal axis 1/T.

Chet

 

[cycrups]

I don't see 1/T on your graph. So this is not correct. Make the horizontal axis 1/T.

Chet

So 1/T would be first changing them to K and then 1 by each of the Temps?

I believe now it's correct http://imgur.com/BkHWdeJ

 

[Chestermiller]

Good. Now go back to the equation you gave in your original post, and take the natural log of the equation. What do you get?

Chet

 

[cycrups]

Good. Now go back to the equation you gave in your original post, and take the natural log of the equation. What do you get?

Chet

Do I multiply each side by the natural log?

but if I multiply the natural log on the left side I will end up with Ln a * b/T. So the question is should I just multiply the left side and assume the equation is = 0?

 

[Chestermiller]

Huh?

What is the natural log of the left side of the equation?
What is the natural log of the right side of the equation?
If the left side is equal to the right side, does that mean that the log of the left side is equal to the log of the right side?

Chet

 

[cycrups]

Huh?

What is the natural log of the left side of the equation?
What is the natural log of the right side of the equation?
If the left side is equal to the right side, does that mean that the log of the left side is equal to the log of the right side?

Chet

Natural log of the left side is → (I think it's like this, but please correct me I'm having a hard time doing this) ln R * ln 1/T
Natural log of the right side is → ln a * 1/T

 

[Chestermiller]

Natural log of the left side is → (I think it's like this, but please correct me I'm having a hard time doing this) ln R * ln 1/T

The R(1/T) on the left side of the equation does not mean that R is being multiplied by 1/T. It is saying that R is a function of 1/T. The natural log of the left side is therefore just ln R

Natural log of the right side is → ln a * 1/T

This is incorrect. The natural log of the right side is ln a + b (1/T)
Is this a physics course, or is it a math course in which you are learning logarithms for the first time? In either case, you need to go back and review logarithms.

Chet

 

[cycrups]

The R(1/T) on the left side of the equation does not mean that R is being multiplied by 1/T. It is saying that R is a function of 1/T. The natural log of the left side is therefore just ln R

This is incorrect. The natural log of the right side is ln a + b (1/T)
Is this a physics course, or is it a math course in which you are learning logarithms for the first time? In either case, you need to go back and review logarithms.

Chet

just one question why am I in the first place taking the natural log of both side is it to try to eliminate one of the parameters from a and b?

 

[Chestermiller]

just one question why am I in the first place taking the natural log of both side is it to try to eliminate one of the parameters from a and b?

No. When you plotted the data for lnR vs 1/T, it came out to a straight line. You can determine the value of the parameters a and b in your equation by getting the slope and intercept of that straight line.

Chet

 

[cycrups]

No. When you plotted the data for lnR vs 1/T, it came out to a straight line. You can determine the value of the parameters a and b in your equation by getting the slope and intercept of that straight line.

Chet

Thanks a lot I have got the answer!

 

[cycrups]

After you get your x-axis straightened out you can imagine the that the function $a\cdot e^{\frac{b}{T}}$ is a model that describes how your graph works. It can be extended in both directions. If you go to the right $\frac{1}{T}$ increases - which in turn corresponds to a decrease in T, and vice versa. Remember, your y-axis is now the ln(R), where negative values mean R less than 1 and positive values mean R higher than 1 - R will never go negative ;)..

You will observe that if you extend a line from your graph to the left, corresponding to a temperature increase, it will eventually intersect with x- and y-axis. Imagine that your temperature increases to infinity $T \rightarrow \infty$, this means that $\frac{1}{T}\rightarrow 0$ (You will of course never reach this point). What value will $a\cdot e^{\frac{b}{T}}$ have then? This will solve one of your variables.

Further if your line crosses the x-axis which is situated at y=0 (which corresponds to R=1), what value must your other variable have then? ;)
Hint: Solve $R = a\cdot e^{\frac{b}{T}}$ where you now only have 1 unknown variable.

So first I get the slope which is negative because the line is going downwards. Let's assume the slope is -12.66 what do I do after this step? I know that I need to multiply ln by both sides of the gives equation so I will get
lnR = a*(b/t)
How do I find a and b from the equation if both of them are unkown and how can I use the slope to solve the problem for a and b?

 

[Chestermiller]

So first I get the slope which is negative because the line is going downwards. Let's assume the slope is -12.66 what do I do after this step? I know that I need to multiply ln by both sides of the gives equation so I will get
lnR = a*(b/t)
How do I find a and b from the equation if both of them are unkown and how can I use the slope to solve the problem for a and b?

You need to review how to work with logarithms.

lnR = ln a + b(1/T)

Chet

 

[cycrups]

You need to review how to work with logarithms.

lnR = ln a + b(1/T)

Chet

Ah yes sorry forgot to put the Ln a on there had it while doing the work on paper. What's the next step how I can get to find the parameters a and b? My Prof hinted b should be the slope but I am not sure if I heard him correctly. I tried to figure it out and find out how b can be the slope but II couldn't find no answer.

 

[Chestermiller]

Ah yes sorry forgot to put the Ln a on there had it while doing the work on paper. What's the next step how I can get to find the parameters a and b? My Prof hinted b should be the slope but I am not sure if I heard him correctly. I tried to figure it out and find out how b can be the slope but II couldn't find no answer.

Suppose you substituted y = lnR, and x = (1/T). Then you would have:

y = b x + ln a

What would be the slope of this straight line, and what would be its intercept?

Chet

 

[cycrups]

Suppose you substituted y = lnR, and x = (1/T). Then you would have:

y = b x + ln a

What would be the slope of this straight line, and what would be its intercept?

Chet

b(slope) =
0.0002368421

and by taking y and x from the plotted points on my graph y= 3.85 AND x = 3*10^-3

3.85 = (0.0002368421 * 3.10^-3) + ln a => exp ^ [(3.85)-(0.0002368421*(3*10^-3)] = a

one question how did you get the formula y = bx + ln a

 

[Chestermiller]

b(slope) =
0.0002368421

and by taking y and x from the plotted points on my graph y= 3.85 AND x = 3*10^-3

3.85 = (0.0002368421 * 3.10^-3) + ln a => exp ^ [(3.85)-(0.0002368421*(3*10^-3)] = a

one question how did you get the formula y = bx + ln a

As I said, I just made the substitutions that I indicated.

Chet

 

[Chestermiller]

Please show the details of how you calculated that slope.

Chet