Question #347242

A platinum resistance thermometer consists of a coil of 0.10-mm-diameter platinum wire wrapped in a coil. (a) What is the length of the wire needed so that the coil’s resistance at 20^{o} C is 25 Ω. The resistivity of platinum at this temperature is 1.0 x 10^{-7} Ω.m. (b) What is the temperature of this platinum wire when its resistance increases 35% of its initial temperature? The temperature coefficient of resistivity for platinum is 3.92 x 10^{-3} /^{o} C.

Expert's answer

Given:

"d"=0.1 mm=0.0001 m;

"t1"=20^{0 }C;

"R1"=25 Ω, resistance at 20^{0 }C temperature;

"R0"=1.0 x 10^{-7} Ω.m, resistivity at 20^{0 }C temperature;

"\\alpha"=3.92 x 10^{-3} /^{o} C, temperature coefficient of resistivity for platinum;

"R\/R1=1.35;" here R- resistance of platinum wire at the final temperature;

Find:

a) length of the wire;

b) temperature of the wire;

Solution:

a) "R1=R0(L\/A); \nL=(R1\/R0)A;\nA=\n\\pi(d\/2)^2;"

"L=(R1\/R0)\\pi(d\/2)^2=(25\\Omega\/10^(-7) \u03a9.m)*3.14*((0.0001m)\/2)^2=1.96m."

b) "R=R1(1+\\alpha(t-t1));" "t=(1\/\\alpha)*(R\/R1-1+\\alpha*t1)=(1\/3.92*10^(-3))*(1.35-1+3.92*10^(-3)*20)= 109C;"

**Answer:**

**a) L=1.96m;**

**b) t=109**^{0}**C. **

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