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Thermometers- calculate the centigrade temperature

  • #1

Homework Statement



a constant volume gas thermometer was calibrated at the ice point and the steam point,. the thermometer was there used to measure the temperature of a beaker of hot water, the readings fro the height difference. h, between the mercury level in the closed limp and the open limp were - 50mm for ice point +220mm for steam point and +105 for the temperature of the hot water .

calculate the centigrade temperature of the hot water.

Homework Equations



attachment

The Attempt at a Solution



after applied all the data which was gave into the equation, 41.67 C, was the answer. what am i suppose to do after ? it is the right attempt in the first place?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Attachments

Answers and Replies

  • #2
208
0
The formula is right and i am getting 57.4 C
 
  • #3
The formula is right and i am getting 57.4 C
first of all .. thank you :)!
is that the finally answer ? they are asking for centigrade temperature of the hot water...
what am i suppose to do ? the height / length of the hot water is +105 mm
 
  • #4
208
0
you are welcome......
First tell me what you did....how did you get 41.67....them maybe i can start helping you......
Bye the way ya.....my final answer is 57.4.....ie temp of hot water.....is it correct?
 
  • #5
you are welcome......
First tell me what you did....how did you get 41.67....them maybe i can start helping you......
Bye the way ya.....my final answer is 57.4.....ie temp of hot water.....is it correct?

thanks again
based on the formula..

-50 - 0 / 100 - 220 * 100 C = 0.4167 * 100 = 41.67 C
( -50 is ice point, + 220 is steam point)

and a specific length is given for " temp of hot water" which is +105mm
therefore i doubt that 57.4 c = temp of hot water while +105mm wasnt used.
 
  • #6
208
0
Hold on....let me confirm what we want ......here we want the temperature of hot water in some scale where the ice point is -50 and steam point is 220 and the reading for the hot water is 105 right?

if that is what you are looking at i dont thin you have applied the formula correctly.....
 
  • #7
i have to be honestly with you, i;m currently working on some past paper, which my teacher didn't teach us anything about this topic. but dash the formula on the board. however, the question is " calculate the centigrate temperture of the hot water"
i do not have an idea of what i'm suppose to do. i was just tryingggg
 
  • #8
208
0
Its okay...i will try to help you......
The question is not 100percent clear to me.... the language....but these are the sort of standard questions where the give you the reading of the ice and the steam points and the reading at some other unknown temperature and they tell you to find the temperature.....

let me explain the formula......

theta=((x at theta - x at 0)/(x at 100C - x at 0C))*100C......here x can be any physical quantity be it length of mercury column or whatever.....now you substitute the required values of the physical quanties and you get your answer.....

This is all i can explain here...If you understood good.....or else your teacher has not taught you this right? if she teaches you and explains it you will get it....if you want it now you can refer to your text book.....if they have some explanation......

If you understood what i have told you then try and apply the formula right now.....
 
  • #9
i do understand the fact that x can be any physical quantity, and that is exactly why i inserted -50 and +220 into the formula.
theta = (( -50 - 0) - ( +220 - * what goes here* )) * 100 c
 
  • #10
208
0
dont you think it should be
theta= (x at theta -(-50))/(+220-0))*100.....

I have a surprise for you there is a mistake in what i have written above you have to find it out......i have made intentionally only one mistake(the rest is correct)......so try and understand where i am wrong and tell me.....this will develop your thinking......

please realise where you have made your mistake too.....if you say (-50 -0) in what sense are you writing it......tally your equation with what i told you......

If you dont get it i will tell you....but first understand.....
 
  • #11
(x at 100C - x at 0C) <---- it should be x at theta
right ?
theta= (x at theta -(-50))/(+220-0))*100

what is " x at theta " ?
 
  • #12
208
0
(x at 100C - x at 0C) <---- it should be x at theta
right ?
I did not get which part you are refering to.....And what is my mistake?
 
  • #13
208
0
Better quote my equation and point out my mistake....
 
  • #14
theta=((x at theta - x at 0)/(x at 100C - x at 0C))*100C.....
where x at 0 C should be x at theta instead....
 
  • #15
208
0
wait...are you refering to post no 8? That is correct because i was teaching you the thing there.......No mistake there......the mistake is in post 10.....
 
  • #16
wow i really dont see it -.-!!! sorry
 
  • #17
208
0
theta=((x at theta -(-50))/(+220-0))*100.....
It should be ((x at theta -(-50)/(+220-x at zero))*100....

x at theta= 105 and x at zero as you know is -50.....Try and understand this.....and if you put in the vaules you get 57.4...i guess.....
 
  • #18
it makes alot sense now.. but the equation states taht
theta = (( x at theta - x at 0)/ ( x at 100C - x at theta)) * 100 C
instead of
theta=((x at theta - x at 0)/(x at 100C - x at 0C))*100C

and according to you .... ( assumption)
x at theta = + 105mm
therefore theta = (( 105-50)/ ( 220- 105 )) * 100 C
theta = 47.8 C
 

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