The electrical resistance in ohms of a cerbain thermomter varies with temperature according to `R=R_(0){1+5xx10^(-3)(T-T_(0))}`. The resistance is `101.6Omega` at the triple point of water, and `185.5Omega` at the normal melting point of lead (600.5K). What is the temperature when the resistance is `123.4Omega`?

Explanation

Given `R=R_(0){1+5xx10^(-3)(T-T_(0))}`
`R_(1)=101.6W,T_(1)=273.15K`
`R_(2)=165.5W,T_(2)=600.5K`
`R=123.4W,T=?`
`101.6=R_(0){1+0.005(600.5-T_(0))}" "........(1)`
`165.5=R_(0){1+0.005(600.5-T_(0))}" "..........(2)`
`(1)-:(2)` gives
`1.8258=(1+3.0025-0.005T_(0))/(1+1.365-0.005T_(0))`
i.e. `(4.3193)-(0.00912T_(0))=(4.0025-0.005T_(0))`
`0.3168=0.00412T_(0)`
i.e. `T_(0)=76.89K`
Hence `101.6=R_(0){1+0.005(273.15-76.89)}` gives
`101.6=R_(0){1+0.9815}`
`:.R_(0)=51.279Omega`
and `R=51.279{1+0.005(T-76.89)}` gives
`23.4=51.279{1+0.005T-0.38445}`
`2.4064=[0.61555+0.005T]`
`1.79085=0.0057`
or `T=358.17K`.
At 358.17K, the resistance will be `123.4Omega`.