The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R= R_(0)[ 1 + alpha (T - T_(0))] . The resistances is 101.6 Omega at the triple point of water 273.16 K, and 165.5 Omega at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Omega?

The electrical resistance in ohms of a certain thermometer varies with

1.	The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R= R_(0)[ 1 + alpha (T - T_(0))] . The resistances is 101.6 Omega at the triple point of water 273.16 K, and 165.5 Omega at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Omega?
Answer» SOLUTION :Here, `R_(0) = 101.6 OMEGA, T_(0) = 273.16` K Case (i) `R_(1) = 165.5 Omega, T_(1) = 600.5K ` Case (ii) `R_(2) = 123.4 Omega , T_(2) = ?` USING the relation R = `R_(0) [1 + ALPHA (T - T_(0) )]` Case (i) `165.5 = 101.6 [1 + alpha (600.5 - 273.16) ]` `alpha = (165.5 - 101.6 )/(101.6 xx (600.5 - 273.16)) = (63.9)/(101.6 xx 372.34)` Case (ii) 123.4 = 101.6`[ 1 + alpha (T_(2) - 273.16)]` or `123.4 = 101.6 [ 1 + (63.9)/(101.6 xx 327.34) (T_(2) - 273.16)] ` = 101.6 + `(63.9)/(327.34) (T_(2) - 273.16)` or `T_(2) ((124.4 - 101.6)xx 327.34)/(63.9) + 273.16` = 111.67 + 273.16 = 384.83K