$$
begin{equation*}
Delta L = alpha cdot L_i cdot Delta T
end{equation*}
$$

Where $alpha$ is the coefficient of thermal expansion of steel $alpha = 12 cdot 10^{-6} cdot frac{1}{K}$, $L_i$ the initial length of the bridge $L_i = 3910 text{m}$, and $Delta T$ is calculated as follows:

$$
begin{align*}
& Delta T = T_2 – T_1 \
& Delta T = 30 text{textdegree}text{C} – (-5text{textdegree}text{C}) = 35text{textdegree}text{C}
end{align*}
$$

Finally we have:

$$
begin{align*}
Delta L &= alpha cdot L_i cdot Delta T \
Delta L &= 12 cdot 10^{-6} cdot 3910 cdot 35 \
Delta L &= 1.64text{m}
end{align*}
$$

– The length of a meterstick $L_i = 1text{m}$

– The coefficient of thermal expansion of aluminum $alpha = 24 cdot 10^{-6} cdot frac{1}{text{textdegree}text{C}}$

– The change of length $Delta L = 1text{mm} = 10^{-3}text{m}$

We will use the following formula for the change of length:

$$
begin{equation*}
Delta L = alpha cdot L_i cdot Delta T
end{equation*}
$$

It follows:

$$
begin{equation*}
Delta T = frac{Delta L}{(alpha cdot L_i)}
end{equation*}
$$

Inserting the values we get:

$$
begin{align*}
Delta T &= frac{(1 cdot 10^{-3}text{m})}{(24 cdot 10^{-6} cdot frac{1}{text{textdegree}text{C}} cdot 1text{m})} \
Delta T &= 41.7text{textdegree}text{C}
end{align*}
$$