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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 ? |
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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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