question_answer 3)

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law \[{{T}_{F}}'-{{T}_{F}}=\frac{180}{100}\times 1=\frac{9}{5}\] The resistances is \[273\cdot 16\] at the triple-point of water 273.16 K, and \[1\cdot 250\times {{10}^{5}}Pa\] at the normal melting point of lead (600.5 K). What is the temperature when the resistance is \[1\cdot 797\times {{10}^{5}}Pa\] ?

Answer:

Here \[0\cdot 287\times {{10}^{5}}pa\] \[{{T}_{tr}}=273\cdot 16K.\] Case (i) \[\text{T=}\frac{\text{P}}{{{\text{P}}_{\text{tr}}}}\text{ }\!\!\times\!\!\text{ 27}3\cdot 16=\frac{1\cdot 797\times {{10}^{5}}}{1\cdot 250\times {{10}^{5}}}\times 273\cdot 16=392\cdot 69K\] \[\text{T=}\frac{\text{P}}{{{\text{P}}_{\text{tr}}}}\text{ }\!\!\times\!\!\text{ 27}3\cdot 16=\frac{0\cdot 287\times {{10}^{5}}}{0\cdot 200\times {{10}^{5}}}\times 273\cdot 16=391\cdot 98K\] Case (ii) \[\cdot \] \[\cdot \] Using the relation \[\cdot \] case (i) \[=1\cdot 20\times {{10}^{-5}}\] \[^{-1}\] \[{{45}^{o}}C\] Case (ii) \[L'=L+\vartriangle L=L+\alpha L\vartriangle T\] Or \[100+\left( 1\cdot 20\times {{10}^{-5}} \right)\times 100\times \left( {{45}^{o}}-{{27}^{o}} \right)=100\cdot 0216cm.\]\[{{45}^{o}}C=100\cdot 0216/100\] OR \[{{45}^{o}}C=100\cdot 0216/100\]\[\vartriangle V=\left( V+\vartriangle V \right)-\] \[V=100\cdot 5\]