The folliwing observation were recorded on a platinum resistance thermometer. Resistance at melting point of pressure `=4.71 Omega,` and resistance at `t^(@)C=2.29Omega.` Calculate
Temperature coefficient of resistants of platinum.
Value of temperature t.

Explanation

Given, resistance at melting point of ice, `R_(o)=3.70Omega` Resistance at boiling point of water at normal pressure,
`R_(100)=4.71Omega`
Resitance at `t^(@)C,R_(t) = 5.29 Omega`
According of the formuls, temperature coefficient of resistance is given by
`alpha=(R_(100)-R_(o))/(R_(o)xx100)=(4.71-3.70)/(3.70xx100)=(1.01)/(370)=2.73xx10^(3)//^(@)C`
According to the formula, for temperature t, we have `t=100^(@)Cxx(R_(t)-R_(o))/(R_(100)-R_(o))`
`100^(@)Cxx(5.29-370)/(4.71-3.70)=100^(@)Cxx(1.59)/(1.01)=157.4^(@)C`