A platinum resistancethermometer reads `0^0`at the ice point and boiling point of water respectively. The resistance of a platnum wire varies with Celsius temperature `theta` as `R_t = R_0 (1+alphatheta + betatheta^2)`, where `alpha = 3.8 xx 10^(-3) ^0 C^(-1)` and `beta= -5.6 xx 10^(-7) ^0 C^(-2)`. What will be the reading of this thermometer if it is pleaced in a liquid bath maintained at `50^0 C`?

Answer

`50.4^@C`

Explanation

The resistances of the wire in the thermometer at `100^0 C` and `50^0 C` are
`R_100 = R_0 [1 + alpha xx 100^0 C + beta xx (100^0 C)^2 ]
and, ` R_50 = R_0 [ 1 + alpha xx 50^0 C + beta xx (50^0 C)^2]`.
The temperature `t` measured on the platinum thermometer is given by
`t = (R_50 - R_0)/(R_100 - R_0) xx 100^0`
= (alpha xx 50^0 C + beta xx (50^0 C)^2)/(alpha xx 100^0 C + beta xx (100^0 C)^2 ) xx 100^0`
`=50.4^0`.