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

By the definition of the linear thermal expansion we have:

$$\dfrac{\Delta L}{L_0} = \alpha \Delta T = \alpha (T_{final} - T_{initial}) ,$$

here, $\Delta L = L - L_0$ is the increase in length of the stainless steel probe when it is heated in the furnace, $L_0 = 0.1622 m$ is the initial length of the stainless steel probe at $T_{initial} = 20 \ ^{\circ}C$, $L = 0.1641 m$ is the length of the stainless steel probe at temperature $T_{final}$, $\alpha = 16.0 \cdot 10^{-6} \ ^{\circ}C^{-1}$ is the linear expansion coefficient for steel, $\Delta T$ is the change in temperature, $T_{initial} = 20 \ ^{\circ}C$ is the initial temperature of the stainless steel probe, $T_{final}$ is the final temperature of the stainless steel probe (or the temperature of the furnace).

Then, from this formula we can find the temperature of the furnace:

$$T_{final} = \dfrac{L - L_0}{\alpha L_0} + T_{initial},$$ $$T_{final} = \dfrac{0.1641m - 0.1622m}{16.0 \cdot 10^{-6} \ ^{\circ}C^{-1} \cdot 0.1622m} + 20 \ ^{\circ}C = 752 \ ^{\circ}C.$$

Answer:

$T_{final} = 752 \ ^{\circ}C.$