Bimetallic strip

The result also depends on the elastic strength (Young's modulus) of the materials. If one of the materials is very stiff and the other one is very soft then there will no curvature.

Nevertheless, for the materials in this example the answer will be still in the correct order of magnitude.

By increasing the temperature by the value $\Delta T$ , the metals expand, resulting in new lengths of $\begin{aligned} l_\text{Zn} &= l_0(1 + \alpha_\text{Zn} \Delta T )\\ l_\text{Fe} &= l_0 (1 + \alpha_\text{Fe} \Delta T ) \end{aligned}$ where $\alpha = \frac{1}{l} \frac{\Delta l}{\Delta T}$ is the thermal expansion coefficient.

The metal strips each have the shape of a circular arc with the arc lengths $\begin{aligned} l_\text{Zn} &= \left(r + \frac{d}{2} \right) \phi \\ l_\text{Fe} &= \left(r - \frac{d}{2} \right) \phi \end{aligned}$ with the radius of curvature $r$ of the bimetallic strip and the angle $\phi$ of the circle segment. Thus, the length ratio yields $\begin{aligned} & & \frac{l_\text{Zn}}{I_\text{Fe}} = \frac{r + \frac{d}{2}}{r - \frac{d}{2}} &= \frac{1 + \alpha_\text{Zn} \Delta T }{1 + \alpha_\text{Fe} \Delta T } \\ \Rightarrow & & \left(r + \frac{d}{2}\right) (1 + \alpha_\text{Fe} \Delta T ) &= \left(r - \frac{d}{2}\right) (1 + \alpha_\text{Zn} \Delta T ) \\ \Rightarrow & & r (\alpha_\text{Fe} - \alpha_\text{Zn}) \Delta T &= -\frac{d}{2} \left(2 + (\alpha_\text{Fe} + \alpha_\text{Zn}) \Delta T\right) \approx -d \\ \Rightarrow & & r &\approx \frac{d}{(\alpha_\text{Zn} - \alpha_\text{Fe}) \Delta T} \\ && &= \frac{2 \cdot 10^{-3} }{2 \cdot 10^{-5} \cdot 10^2} \,\text{m} = 1 \,\text{m} \end{aligned}$