Converting between a known and unknown temperature scale

[BOAS]

Homework Statement

On a Temperature scale, the ice point is 25.6°I and the steam point is 155°I. If the temperature reads 66.6°I what is the equivalent in °C?

The Attempt at a Solution

25.6°I = 0°C

155°I = 100°C

1°I = 5/647 of the interval between boiling and freezing.

1C° = 647/500 I°

1I° = 500/647C°

66.6°I = 51.5°C

I arrive at this answer but I'm not convinced I've done this correctly. When changing from faranheit to celsius for example, I have to subtract the offset from zero...

So should my answer actually be 51.5 - 25.6 °C?

I'd really appreciate some help,

thanks!

 

[DrClaude]

66.6°I = 51.5°C

That's not correct. And I don't understand where you get fractions like 5/647.

I think that an easier way to do the problem is to try to solve for the formula
$$
a T_I + b = T_C,
$$
where $T_I$ is the temperature in °I and $T_C$ is the temperature in °C. Using the data you have, that gives you two equations for two unknowns ($a$ and $b$).

 

[DrClaude]

After recalculating a few things, I find that

1C° = 647/500 I°

1I° = 500/647C°

is correct. The problem is the shifting of the scale. Neither

66.6°I = 51.5°C

nor

So should my answer actually be 51.5 - 25.6 °C?

are correct.

 

[HallsofIvy]

Note that Dr. Claude is using the fact that the relation between the two temperature scales is linear- one degree celsius converts to the same number of degrees Farenheit, and vice-versa, at any temperature.

 

[BOAS]

That's not correct. And I don't understand where you get fractions like 5/647.

I think that an easier way to do the problem is to try to solve for the formula
$$
a T_I + b = T_C,
$$
where $T_I$ is the temperature in °I and $T_C$ is the temperature in °C. Using the data you have, that gives you two equations for two unknowns ($a$ and $b$).

When converting between faranheit and celsius, we say that 1°F = 1/180 of the interval between the boiling and freezing point.

Compared to the 1/100 of the interval that 1°C expresses.

9/5 of a degree faranheit is equivalent to 1 degrees celsius.

It was this method I was trying to use that gave me those strange fractions...

I haven't seen the method you are talking about before, please could you apply it to something so I can see what you mean?

 

[DrClaude]

Note that Dr. Claude is using the fact that the relation between the two temperature scales is linear- one degree celsius converts to the same number of degrees Farenheit, and vice-versa, at any temperature.

Right. It is implicit in the problem. Good of you for pointing this out.

 

[BOAS]

After recalculating a few things, I find that

is correct. The problem is the shifting of the scale. Neither

nor

are correct.

Ok,

I think this makes sense to me, I hope it does to you.

°C = ([°I] - 25.6) x 500/647

giving me the answer that 66.6°I = 31.7°C

 

[DrClaude]

Edit: You got the correct answer in the previous post while I was typing the following. I'll leave it in case someone else has a similar problem.

I haven't seen the method you are talking about before, please could you apply it to something so I can see what you mean?

Lets say we want to rederive the conversion from Fahrenheit to celsius. Take

freezing point: 32 °F = 0 °C
boiling point: 212 °F = 100 °C

We look for an equation of the type
$$
a T_F + b = T_C
$$
For the values above
$$
a \times 32 + b = 0 \\
a \times 212 + b = 100
$$
From the first equation, we have $b = -32 a$. Using that in the second equation, we get
$$
212 a - 32 a = 100 \Rightarrow a = \frac{100}{180} = \frac{5}{9}
$$
and $b = -160 / 9$. So
$$
\frac{5}{9} T_F - \frac{160}{9} = T_C
$$

I realize now that this is not the usual formula. You should therefore try it yourself using
$$
\left( T_F + a \right) \times b = T_C
$$
to get the factors you know. You can then do the same for your problem.

 

[DrClaude]

Ok,

I think this makes sense to me, I hope it does to you.

°C = ([°I] - 25.6) x 500/647

giving me the answer that 66.6°I = 31.7°C

Correct!

 

[BOAS]

Correct!

Thank you for the help

 

[Iamtheonewhoknocks]

This is a detailed explanation on how to solve this problem and solve any problem based on temperature scale conversion. It's very easy. Hope it helps. I AM THE ONE WHO KNOCKS.