1.

What is the temperature of the triple - point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale ?

Answer»

Solution :Rotation between Fahrenheit and kelvin scale.
`(T_(F)-32)/(180)=(T_(K)-273)/(100)`. . . (1)
FO different TEMPERATURE,
`(T_(F)^(.)-32)/(180)=(T_(K)^(.)-273)/(100)`. . . (2)
By subtracting equating (2) from equation (1)
`(T_(F)^(.)-T_(F))/(180)=(T_(K)^(.)-T_(K))=(T_(K)^(.)-T_(K))/(100)`
`:.T_(F)^(.)-T_(F)=(180)/(100)(T_(K)^(.)-T_(K))`
But `T_(K)^(.)-T_(K)=1K`
`:.T_(F)^(.)-T_(F)=(180)/(100)`
Now new temperature at triple point is `273.16K`,
Tmeprature `=273.16xx(180)/(100)`
`=491.688" "^(@)F~~491.69^(@)F`


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