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The electrical resistance in ohms of a c...

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law
`R=R_(0)[1+a(T-T_(0))]`
The resistance 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`?

Text Solution

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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
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