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The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: 

                  R Ro [1 + α (– To)] 

The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?

Given that, 


straight R space equals space straight R subscript straight o space left square bracket space 1 space plus space straight alpha space left parenthesis straight T minus straight T subscript straight o right parenthesis right square bracket space space space space space space space space space space space space space... space left parenthesis straight i right parenthesis thin space

where comma space

straight R subscript straight o space end subscript is space the space initial space resistance comma space and space

straight T subscript straight o space is space the space initial space temperature comma space

straight R space is space the space final space resistance comma space and

straight T space is space the space finl space temperature. space

At space triple space point space of space water comma space straight T subscript straight o space equals space 273.15 space straight K space

Resistance space of space lead comma space straight R subscript straight o space equals space 101.6 space straight capital omega

At space normal space melting space point space of space lead comma space straight T space equals space 600.5 space straight K

Resistance space of space lead comma space straight R space equals space 165. space 5 space straight capital omega space

Putting space values space in space equation space left parenthesis straight i right parenthesis comma space we space have

165.5 space equals space equals space 101.6 space left square bracket space 1 plus straight alpha space left parenthesis 600.5 space minus space 273.15 right parenthesis right square bracket space

1.629 space equals space 1 space plus space straight alpha space left parenthesis 327.35 right parenthesis space

therefore space straight alpha space equals space fraction numerator 0.629 over denominator 327.35 end fraction space equals space 1.92 space cross times space 10 to the power of negative 3 end exponent space straight K to the power of negative 1 end exponent space

For space Resistance comma space straight R subscript 1 space equals space 123.4 space straight capital omega space comma space we space have

123.4 space equals space 101.6 space left square bracket 1 space plus space 1.92 space cross times 10 to the power of negative 3 end exponent space left parenthesis straight T space minus space 273.15 right parenthesis right square bracket

On space solving space for space straight T comma space we space get

straight T space equals space 384.61 space straight K comma space is space the space temperature space when space resistance space is space 123.4 space straight capital omega.

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The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?

The triple point of water has a unique value of 273.16 K. At varying values of volume and pressure, the triple point of water is always 273.16 K.

The melting point of ice and boiling point of water do not have particular values because these points depend on pressure and temperature.
278 Views

The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.


Kelvin and Celsius scales are related as: 

TC = TK – 273.15                  … (i) 

Celsius and Fahrenheit scales are related as: 

TF = (9/5)TC + 32                ...(ii)


For neon: 

TK = 24.57 K 

 T= 24.57 – 273.15 = –248.58°C 

TF = (9/5)TC + 32 

    = (9/5) × (-248.58) +32 

    = 415.440 F

For carbon dioxide:

TK = 216.55 K 

 TC= 216.55 – 273.15

     = –56.60°C 

TF = (9/5)TC + 32 

    = (9/5) × (-56.60) +32 

    = -69.880 C
462 Views

Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation betweenTA and TB?

Triple point of water on absolute scale A, T1 = 200 A

Triple point of water on absolute scale B, T2 = 350 B

Triple point of water on Kelvin scale, TK = 273.15 K

The temperature 273.15 K on Kelvin scale is equivalent to 200 A on absolute scale A.

           T1 = T

      200 A = 273.15 K 

∴          A = 273.15/200 

The temperature 273.15 K on Kelvin scale is equivalent to 350 B on absolute scale B. 

          T2 = T

     350 B = 273.15

∴        B = 273.15/350 

TA is triple point of water on scale A. 

TB is triple point of water on scale B. 

∴         273.15/200 × T = 273.15/350 × T

Therefore, the ratio TA : Tis given as 4 : 7
267 Views

There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale?

The absolute zero or 0 K is the other fixed point on the Kelvin absolute scale.
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