Extrapolating from a calorimetry lab to find a new delta T with different volumes and grams of substance
The ratio strategy will definitely work, but I think going back to $q=m\cdot C\cdot \Delta T$ is easier to understand.

You've already figured out a bunch of these constants: $m = 50$, $C = 4.184$ (we assume that the heat capacity of the water doesn't change too much when adding the solid), and $\Delta T = 13.33$.

$$\begin{align}
q &amp; =m\cdot C\cdot \Delta T \\
&amp; = 50\cdot4.184\cdot13.33 \\
&amp; = \ce{2789 J}
\end{align}$$

The heat produced by mixing $\ce{5 g}$ of $\ce{CaCl_2}$ is $\ce{2789 J}$, therefore,
the heat of reaction ($\ce{\Delta H_{rxn}}$) in $\ce{J/g}$ is $\ce{557.7 J/g}$.

When we mix $\ce{10 g}$ of $\ce{CaCl_2}$, the total heat produced will be $\ce{10 g \cdot 557.7 J/g = 5577 J}$.

Now we need to determine the heat change had we mixed in $\ce{10 g}$ of $\ce{CaCl_2}$. We can rearrange the specific heat formula to get $\ce{\Delta T = \frac{q}{m\cdot C}}$. We know $q = 5577$, $m=50$, and $C=4.184$.

$$\begin{align}
\Delta T &amp;= \frac{q}{m\cdot C} \\ &amp; \\
&amp;=\frac{5577}{50\cdot4.184} \\ &amp; \\
&amp;= 26.66\, ^\circ \ce{C}
\end{align}$$

The reason the ratio works is the same logic as the gas laws:

$P_1\ V_1 = C$, $P_2\ V_2 = C$, therefore $P_1\ V_1 = P_2\ V_2$.

$q_1 / T_1 = m\cdot C$ and $q_2 / T_2 = m\cdot C$.

Therefore, $q_1 / T_1 = q_2 / T_2$ by the transitive property of equality