Hannah M.

asked • 08/27/16

Convert 2316 oV to oM

I can go on the internet to find these conversions but I am hoping that someone will be able to show the steps to do this. Specifically, I am supposed to convert between temperature scales by using an appropriate conversion factor after normalizing common points of reference.
 
1. Convert 2316 oV to oM
 
2. Convert 300 K to oF
 
3. Convert 48230 oL to oC
 
Thank you in advance!

Kendra F.

Some of these temperatures I've never seen of heard of before as a chemistry major.
There are probably formulas for converting between these temperature units.
 
Knowing that;
K = C + 273.15
C = 5/9( F - 32)

substituting that Eq. in for C will give you a conversion for F to K or vice versa
 
 
K = 5/9(F - 32) + 273.15

(K - 273.15)(9/5) + 32 = F
 
(300 - 273.15)(9/5) + 32 = F
 
26.85(9/5) + 32 = F
 
48.3 + 32 = F
 
80.3 ºF = 300 K
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08/27/16

Steven W.

tutor
Hi Hannah!  I might just be out of practice, but I do not recognize V, M, or L as denoting temperature scales.  F(ahrenheit), C(elsus), and K(elvin), I am familiar with.  But the others are a mystery to me, unless they are defined in your class materials some other way.  Do you have any more information about them?
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08/27/16

Hannah M.

Thank you. I am supposed to follow certain steps, and I am making a mistake(s) somewhere.

Here are the instructions:

"Converting between temperature scales can be done by using an appropriate conversion factor after normalizing common points of reference.

In this exercise, the common points of reference for actual and hypothetical temperature scales listed ate the freezing and boiling points of water. Normalization will be done by simply adding or subtracting degrees to adjust the freezing point to zero. The relevant conversion factor will be comprised of the number of degrees between freezing and boiling points for the two scales of interest as numerator and denominator, or vice versa.

For the following problems, indicate the number to add or subtract to normalize the first temperature scale to zero on line a. List the adjusted temperature on line b. List the appropriate conversion factor on line d. Indicate the adjusted temperature after using the conversion factor on line d. Indicate the number to add or subtract to denormalize the new temperature to the new temperature scale on line e. Finally, List the fully converted temperature on line f.

(K) Kelvin 273 to 373

(F) Fahrenheit 32 to 212"

Question 2. Convert 300 K to F

a. -273 (I did this by 273 + X = 0; X = -273)

b. 27 (I did this by 300 - 273 = 27)

c. 3.70 (I did this by 100/180 = 27/X)

d. 48.65 (I did this by finding X from above)

e. -32 (I did this by 32 + X = ); X = -32

f. 16.65 (I did this by 48.65 + (-32) = 16.65)

I know this is wrong because the online converter says the answer is 80.33 Can you tell where my mistakes are? Thank you!
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08/27/16

Hannah M.

Sorry. My teacher gave me actual and hypothetical temperature scales. Can you see what I posted under the comment from Kendra F? If not, I can post under the comment section below your name as well. Thanks!
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08/27/16

1 Expert Answer

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Kendra F. answered • 08/27/16

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Hi Hannah,

As a Chem major, I've never seen some of the above temperature units. There are most likely formulas for converting these. You could use a reference point then a scaling factor if you can find these online or in your textbook, notes, etc.

K = C + 273.15
 
C = (5/9)(F - 32)
 
if you substitute the above Eq. into the one for Kelvin then you have a conversion for F --> K
 
K = (5/9)(F - 32) + 273.15
 
(K - 273.15)(9/5) + 32 = F 
 
(300 - 273.15)(9/5) + 32 = F
 
(26.85)(9/5) + 32 = F
 
48.33 + 32 = F
 
80.33 ºF = 300 K
 
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Hannah M.

Thanks. I have been working out the conversions but I seem to be getting the wrong answers. I am supposed to do the conversions a certain way.
 
"Converting between temperature scales can be done by using an appropriate conversion factor after normalizing common points of reference.
 
In this exercise, the common points of reference for actual and hypothetical temperature scales listed are the freezing and boiling points of water. Normalization will be done by simply adding or subtracting degrees to adjust the freezing point to zero. The relevant conversion factor will be comprised of the number of degrees between freezing and boiling points for the two scales if interest as numerator and denominator, or visa versa.
 
For the following problems, indicate the number to add of subtract to normalize the first temperature scale to zero on line a. List the adjusted temperature on line b. List the appropriate conversion factor on line c. Indicate the adjusted temperature after using the conversion factor on line d. Indicate the number to add or subtract to denormalize the new temperature scale on line e. Finally, list the fully converted temperature on line f.
 
Scales
 
(K) Kelvin 273 to 373
 
(F) Fahrenheit 32 to 212"
 
Here is what I did for Question 2.
 
2. Convert 300 K to Fahrenheit
 
a      -273 (I got this by looking at the scale for Kelvin and doing this 273 + X = 0; X = -273)
 
b      27 (I got this by 300 - 273 = 27)
 
c      100/180 = 27/X (X is the adjusted temperature)
 
d      666
 
e     -32 (I got this by looking at the scale for Fahrenheit and doing this 32 + X = 0; X = -32)
 
f      634 (I got this by 666 - 32 = 634) 
 
However, I know I did something wrong (but I don't know what) because when I use the online converter, I get 80.33 as the answer. Can you tell where my mistake is? Thank you so much!
 
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08/27/16

Hannah M.

Thank you. I am supposed to follow certain steps, and I am making a mistake(s) somewhere.
 
Here are the instructions:
 
"Converting between temperature scales can be done by using an appropriate conversion factor after normalizing common points of reference.
 
In this exercise, the common points of reference for actual and hypothetical temperature scales listed ate the freezing and boiling points of water. Normalization will be done by simply adding or subtracting degrees to adjust the freezing point to zero. The relevant conversion factor will be comprised of the number of degrees between freezing and boiling points for the two scales of interest as numerator and denominator, or vice versa.
 
For the following problems, indicate the number to add or subtract to normalize the first temperature scale to zero on line a. List the adjusted temperature on line b. List the appropriate conversion factor on line d. Indicate the adjusted temperature after using the conversion factor on line d. Indicate the number to add or subtract to denormalize the new temperature to the new temperature scale on line e. Finally, List the fully converted temperature on line f.
 
(K) Kelvin 273 to 373
 
(F) Fahrenheit 32 to 212"
 
Question 2. Convert 300 K to F
 
a. -273 (I did this by 273 + X = 0; X = -273)
 
b. 27 (I did this by 300 - 273 = 27)
 
c. 3.70 (I did this by 100/180 = 27/X)
 
d. 48.65 (I did this by finding X from above)
 
e. -32 (I did this by 32 + X = ); X = -32
 
f. 16.65 (I did this by 48.65 + (-32) = 16.65)
 
I know this is wrong because the online converter says the answer is 80.33 Can you tell where my mistakes are? Thank you!
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08/27/16

Hannah M.

Thank you!
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08/27/16

Hannah M.

Question 2. Convert 300 K to F
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08/27/16

Hannah M.

a. -273 (I did this by 273 + X = 0; X = -273)
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08/27/16

Hannah M.

b. 27 (I did this by 300 - 273 = 27)
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08/27/16

Hannah M.

c. 3.70 (I did this by 100/180 = 27/X)
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08/27/16

Hannah M.

d. 48.65 (I did this by finding X from above)
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08/27/16

Hannah M.

e. -32 (I did this by 32 + X = ); X = -32
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08/27/16

Hannah M.

f. 16.65 (I did this by 48.65 + (-32) = 16.65)
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08/27/16

Hannah M.

I know this is wrong because the online converter says the answer is 80.33 Can you tell where my mistakes are? Thank you!
 
(I tried to type all of this under one comment but when I press "SAVE COMMENT" my comment would be deleted and nothing posted. Sorry about the multiple comment lines.)
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08/27/16

Kendra F.

Add/Subtract degrees to adjust freezing point to zero.
a) K - 273 = temperature difference between freezing point in water
 
Indicate the adjusted temperature
b) 27 Kelvin
 
List the appropriate conversion factor on line
Conversion factor: ΔF/ΔK is going from K --> F 
(K) Kelvin 273 to 373
ΔK = 100
(F) Fahrenheit 32 to 212
ΔF = 180
c) 180/100 = 9/5
 
Indicate the adjusted temperature after using the conversion factor on line
d) (27 K)(9F/5K) = 48.6 F
 
Indicate the number to add or subtract to de-normalize (add F freezing point) the new temperature to the new temperature scale on line
e) 48.6 F + 32F
 
Finally, List the fully converted temperature on line
f) 80.6 F
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08/27/16

Kendra F.

The "K" in part a) is the temperature you need to convert, 300 K
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08/27/16

Hannah M.

Thank you.
 
Why in e is it +32 and not -32?
 
Why in c is it 180/100 and not 100/180?
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08/27/16

Kendra F.

In (c) you are converting units. You have kelvin and you want to convert to ºF so it has to be ΔF/ΔK so that kelvin cancels out and you are left with ºF.
 
(27 K) * (9 ºF/5 K) = 48.6 ºF
If you have the conversion factor the wrong way, the units become K2/ºF which is not correct.
 
In (e) you add the freezing point of water in ºF because you subtracted the kelvin freezing point in step (a).
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08/27/16

Hannah M.

I think I got it now. Thanks!
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08/27/16

Hannah M.

I have been working on the other problems and have a question about how to get (e).
 
If I am to convert 2316 Violet degrees to Muppet degrees, the same way as I did above, what is (e)?
 
Given (V) Violet 2000 to 2800 and (M) Muppet -3 to +9
 
a. -2000
b. 316
c. 12/800 =3/200
d. 316(3/200) = 4.74
e. 3
f. 4.74 + 3 = 7.74
 
I don't understand if (e) is 3 or -3. Thank you again!
 
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08/27/16

Kendra F.

Hi Hannah,

You can use either the freezing point or boiling point as long as you stay consistent and use the same at the end. So if we use the boiling point instead..
 
2316 - 2800 = -484 V
-484 V * (3 M / 200 V) = -7.26 M
 
-7.26 M + 9 = 1.74 M
 
This is the same answer you get when you add -3.
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08/27/16

Kendra F.

Hi Hannah,
 
You'll need to add -3. 
 
You can check this by using the boiling point as your reference instead of the freezing point.
 
2316 V - 2800 V = - -484 V
-484 V * (3 M / 200 V) = -7.26 M
-7.26 M + 9 M = 1.74 M
 
This is the same temperature you get when you use the freezing point and add -3.
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08/27/16

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