Temperature and Thermometers

[Integrator741]

Good Evening,

I am going through some physic questions, hoping that will improve my chances of succeeding in exam, and I found a few questions about temperature and thermometers that I am not too sure.

1. When a particular temperature is measured on scales based on different properties it has a different numerical value on each scale except at certain points, explain why this is so and state at what points the values agree.

2. The resistance of a wire at temperature x degrees of C measure on a standard scale is given by R(resistance at unknown temperature) = R0(1+A(temperature)+10^-3 * A(temperature)^2), where A is a constant. When the thermometer is at a temperature of 50 C on the standard scale, what will be the temperature indicated on the resistance scale?

I had similar questions before, but I was able to solve them using a simple book called "Explaining Physics GCSE Edition". I know that they are not hard, but I can't understand them.

Thank you.

 

[dorke]

1.If by "scales" they mean Celsius and Fahrenheit the answer is -40F=-40C
2. Not so clear: "what will be the temperature indicated on the resistance scale? "

 

[davenn]

1.If by "scales" they mean Celsius and Fahrenheit the answer is -40F=-40C

interesting that you picked that particular temp .... its the only place the 2 scales read the same temp

 

[Harald Kapp]

explain why this is so a

The temperature scales of Fahrenheit and Celsius are related as follows:
F = C(1.8) + 32
For which value of C is F=C? this gives both the value and the explanation.

When the thermometer is at a temperature of 50 C on the standard scale, what will be the temperature indicated on the resistance scale?

When properly calibrated, the resistive thermometer should also indicate 50 °C

 

[Integrator741]

1.If by "scales" they mean Celsius and Fahrenheit the answer is -40F=-40C
2. Not so clear: "what will be the temperature indicated on the resistance scale? "

I can't understand it myself, this is so unclear.. I thought that I could use that formula, but it doesn't really work for me. Apparently the answer is 47.7 C.

 

[Arouse1973]

God, where did you get those questions from. They seem like they were written by someone with no physics experience at all. Please send me a link, I would love to email them.
Adam

 

[Integrator741]

God, where did you get those questions from. They seem like they were written by someone with no physics experience at all. Please send me a link, I would love to email them.
Adam

my lecturer gave them to me.

 

[Arouse1973]

Please tell me the course and give me his email address.
Thanks
Adam

 

[Arouse1973]

Also do you have a copy of the questions? Just in case you have missed something which might explain it in more detail.
Thanks
Adam

 

[Integrator741]

Also do you have a copy of the questions? Just in case you have missed something which might explain it in more detail.
Thanks
Adam

Hold on, I will upload the question sheet.

 

[Arouse1973]

Cool I will have a look later, just doing curry for tea
Adam

 

[Integrator741]

Hold on, I will upload the question sheet.

Question 8,b

 

[Arouse1973]

Um still not clear to me either.
Adam

 

[Integrator741]

Um still not clear to me either.
Adam

I will ask him after holidays. Thanks for your help

 

[Laplace]

it has a different numerical value on each scale except at certain points,

This image contains all the data necessary to answer the question. Try plotting temperatures in K, C, & F on a vertical scale against a horizontal scale of absolute temperature (i.e., the lowest possible temperature is zero). Note where the plotted lines may intersect.

 

[Integrator741]

This image contains all the data necessary to answer the question. Try plotting temperatures in K, C, & F on a vertical scale against a horizontal scale of absolute temperature (i.e., the lowest possible temperature is zero). Note where the plotted lines may intersect.

View attachment 24114

So don't use the formula at all? Sorry I don't really follow.

 

[Harald Kapp]

Question 8,b

The answer is probably the evaluation of the term R(resistance at unknown temperature) = R0(1+A(temperature)+10^-3 * A(temperature)^2)
such that R(50 °C) = R0*X where X is the evaluated value of the parenthesis for temp=50 °C.
Depending on R0 the absolute value will vary for every length of wire.

 

[Merlin3189]

1.If by "scales" they mean Celsius and Fahrenheit the answer is -40F=-40C

1. When a particular temperature is measured on scales based on different properties it has a different numerical value on each scale except at certain points,

So I think they do not mean C and F, nor K nor Reaumur, etc.

Presumably they mean when measured on a gas scale, resistance scale, mercury in glass scale, etc using whatever unit you choose. Then they would agree at the fixed points, because this is where they are calibrated. The numerical values would depend on the scale chosen, but for all types of thermometer, they would read 0 C at freezing point and 100 C at boiling point. (Ok, not the modern definition, but it's what most people are accustomed to.)

In between they do not necessarily vary linearly nor in step. So a resistance thermometer and a mercury in glass thermometer may well show different readings at -40 C and at any other temperature than the calibrated fixed points. (Of course, the differences may or may not be great and if they vary from positive to negative difference or vice versa, then they must agree at other points.)

As for the numerical part of the question, you can work out an expression for the resistance at the fixed points and at the required 50 C temperature. Then the reading of the resistance thermometer is proportional to the change of resistance from the lower fixed point (0 C) to the measured temp (50 C) compared to the change in resistance from the lower FP to the upper FP (100 C). And since we want 100 degrees from LFP to UFP, then this fraction is multiplied by 100.

So Temperature shown by Resistance Thermometer at 50 C = 100 * (change in R from 0 to 50 C) / (change in R from 0 to 100 C)
= 47.72 degrees C

If you had only a resistance thermometer with no other sort of thermometer to compare it with, then this would have to be your scale. When someone invented a mercury in glass thermometer, they would calibrate it against yours and find theirs had discrepancies from your "true" scale and only matched exactly at the fixed points.

If some foreigner came along and said he wanted to use alcohol as the liquid, so please could we use some other fixed points to stop his alcohol boiling away, we'd all calibrate at the new colder points, agree there, but differ slightly elsewhere and argue about whose was the best system.

Finally (or maybe not so finally) some smart Alec physicist would turn up with some yarn about ideal gases, absolute zero, thermodynamic scales and being able to calculate some temperature we'd all agree on using his almost ideal gas thermometer. I'd take one look at the arcane equations, scratch my head and go back to using my hand and saying, "feels warm enough to me".

 

[Integrator741]

So I think they do not mean C and F, nor K nor Reaumur, etc.

Presumably they mean when measured on a gas scale, resistance scale, mercury in glass scale, etc using whatever unit you choose. Then they would agree at the fixed points, because this is where they are calibrated. The numerical values would depend on the scale chosen, but for all types of thermometer, they would read 0 C at freezing point and 100 C at boiling point. (Ok, not the modern definition, but it's what most people are accustomed to.)

In between they do not necessarily vary linearly nor in step. So a resistance thermometer and a mercury in glass thermometer may well show different readings at -40 C and at any other temperature than the calibrated fixed points. (Of course, the differences may or may not be great and if they vary from positive to negative difference or vice versa, then they must agree at other points.)

As for the numerical part of the question, you can work out an expression for the resistance at the fixed points and at the required 50 C temperature. Then the reading of the resistance thermometer is proportional to the change of resistance from the lower fixed point (0 C) to the measured temp (50 C) compared to the change in resistance from the lower FP to the upper FP (100 C). And since we want 100 degrees from LFP to UFP, then this fraction is multiplied by 100.

So Temperature shown by Resistance Thermometer at 50 C = 100 * (change in R from 0 to 50 C) / (change in R from 0 to 100 C)
= 47.72 degrees C

If you had only a resistance thermometer with no other sort of thermometer to compare it with, then this would have to be your scale. When someone invented a mercury in glass thermometer, they would calibrate it against yours and find theirs had discrepancies from your "true" scale and only matched exactly at the fixed points.

If some foreigner came along and said he wanted to use alcohol as the liquid, so please could we use some other fixed points to stop his alcohol boiling away, we'd all calibrate at the new colder points, agree there, but differ slightly elsewhere and argue about whose was the best system.

Finally (or maybe not so finally) some smart Alec physicist would turn up with some yarn about ideal gases, absolute zero, thermodynamic scales and being able to calculate some temperature we'd all agree on using his almost ideal gas thermometer. I'd take one look at the arcane equations, scratch my head and go back to using my hand and saying, "feels warm enough to me".

Amusing and informative - thank you very much.