Answer to Question #117529 in Molecular Physics | Thermodynamics for alexis

Question

A small probe is made of stainless steel and has a length of 164.1 mm when being used to monitor the temperature in a furnace.
When the probe is removed for maintenance and allowed to cool to 20.0°C the length of the probe is measured as 162.2 mm.
What is the temperature of the furnace? (to 3 s.f and in °C)
[αsteel = 16.0×10−6 K−1]

Answer 1

Solution.

$L=164.1mm;$

$L_0=162.2mm;$

$T_0=20^oC=293K;$

$\alpha_L=16.0\sdot10^{-6}K^{-1};$

$\Delta T-?;$

The relative change in linear dimension, which can be considered as a relative deformation, can be recorded:$\epsilon=\dfrac{\Delta L}{L_0}=\alpha_L\sdot\Delta T\implies\Delta T=\dfrac{\Delta L}{L_0\alpha_L};$

$\Delta L=164.1mm-162.2mm=1.9mm;$

$\Delta T=\dfrac{1.9mm}{162.2mm\sdot 16.0\sdot 10^{-6} K^{-1}}=732K;$

$\Delta T=T-T_0; \implies T=\Delta T+T_0;$

$T=732K+293K=1025K=752^oC;$

Answer:$T=752^oC.$