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Answer the following: (a) 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)? (b) There were two fixed points in the originalCelsuis scales as mentioned above which were assigned the numver 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 273.16K. what is the other fixed point on this (Kelvin) scale? (c ) The absolute temperature (Kelvin scale) T is related to the temperature t_(c) on the Celsius scale by t_(c) = T-273.15 Whay do we have 273.15 in this relation, and not 273.16? (d) 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 :(a) This is on ac count of the fact that the triple point of water has a unique value i.e., `273.16K` at a unique point, where exists unique of pressure and volume. On the other hand, the melting point of ice and boiling point of water do not have unique set of values as they cnange with the change in pressure and volume. (b) The other fixed point on the Kelvin absolute scale is the absolute zero itself. (c) On Celsius scale `0^(@)C` corresponds to melting point of ice at normal pressure. The corresponding value of abslute temperature is `273.15K`. the temperature `273.16K` corresponds to the triple point of water. form the given relation, the corresponding value of triple point of water on Celsius scale =`273.16-273.15 = 0.01^(@)C`. (d) We know that Fahrenheight scale and absolute scale are related as `(T_(F)-32)/(180) = (T_(K)-273.15)/(100)` ....(i) For ANOTHER set of TEMERATURE `T_(F)^(')` and `T_(K)^(')` `(T_(F)^(')-32)/(180) = (T_(K)^(')-273.15)/(100)` ....(ii) Subtracting (i) from (ii) , we have ` (T_(F)^(')-T_(F))/(180) = (T_(K)^(')-T_(K))/(100)` or `T_(F)^(') - T_(F) = 180/100 (T_(K)^(")-T_(K))` if `T_(K)^(')-T_(K) =1K` , then, `T_(F)^(') - T_(F) = 180/100xx1=9/5` For a temperature of triple point i.e., `273.16K`, the temperature on the NEW scale is `=273.16xx9/5 = 491.69`. |
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