Question

a.A traditional thermostat regulates temperature with the use of a bimetallic strip. The bimetallic strip consistsof Iron (a = 11.8 x 10-6 °C-4) and Copper (a = 16.5 x 10-6 °C'') that expand at different rates as thetemperature increases. Assuming that both metal strips are at 10 cm, find the difference in lengths betweenthe two metals as the temperature increases from 20°C (room temperature) to 100°C (boiling temperature).

Answer 1

Given,

The coefficient of thermal expansion of iron, α₁=11.8×10⁻⁶ °C⁻¹

The coefficient of thermal expansion of copper, α₂=16.5×10⁻⁶ °C⁻¹

The length of both strips, L=10 cm=0.1 m

The initial temperature, T₁=20 °C

The final temperature, T₂=100 °C

The increase in the length due to change in the temperature is given by,

$\Delta L=\alpha L\Delta T$

Thus, the change in the length of the iron strip is,

$\begin{gathered} \Delta L_1=11.8\times10^{-6}\times0.1\times(100-20) \\ =94.4\times10^{-6}\text{ m} \end{gathered}$

Therefore the new length of the iron strip when the temperature rises to 100 °C is

$\begin{gathered} L_{n1}=L+\Delta L_1 \\ =0.1+94.4\times10^{-6} \\ =0.1000944\text{ m} \end{gathered}$

The change in the length of the copper strip is given by,

$\begin{gathered} \Delta L_2=16.5\times10^{-6}\times0.1\times(100-20) \\ =132\times10^{-6}\text{ m} \end{gathered}$

Thus the new length of the copper strip is,

$\begin{gathered} L_{n2}=\Delta L_2+L \\ =0.1+132\times10^{-6} \\ =0.100132\text{ m} \end{gathered}$

Thus the difference in the new lengths of the two metal strips is

$\begin{gathered} \Delta l=L_{n2}-L_{n1} \\ =0.100132-0.1000944 \\ =37.6\times10^{-6}\text{ m} \end{gathered}$

Thus the change in the length of the two metal strips when the temperature is increased to 100 °C is 37.6×10⁻⁶ m.