Extrapolating from a calorimetry lab to find a new delta T with different volumes and grams of substance

So, we just finished an AP chem lab where the question was basically:

Given the results of the lab were that $$\pu{5 grams}$$ of $$\ce{CaCl2}$$ and $$\pu{45ml}$$ of water produced a $$\pu{13.33 ^\circ C}$$ change, find the new temperature change of $$\pu{10 grams}$$ of $$\ce{CaCl2}$$ in $$\pu{40ml}$$ of water.

Our teacher used a ratio strategy where basically $$\frac{\pu{5 grams}}{\pu{10 grams}}=\frac{\pu{13.33 ^\circ C}}{x}$$ Then, since x turns out to be $$\pu{26.66 ^\circ C}$$; $$\frac{\pu{26.66 ^\circ C}}{45ml}=\frac{x}{50ml}$$ solving for $$x$$, the final temperature change is $$\pu{29.6 ^\circ C}$$.

My question is basically whether this is true, because I find the denominator in the first fraction of the second ratio sketchy - its saying $$\pu{10 grams}$$ for 35 ml of water, but, the $$\pu{26.66 ^\circ C}$$ is derived from finding $$\pu{10 grams}$$ with same amount of water. Thus, shouldn't it be 50, since in the first ratio we only changed amount of grams of solution, so water is still 40 ml, and now $$\ce{CaCl2}$$ is $$\pu{10 grams}$$. $$\pu{40ml + 10 grams = 50 grams}$$. Thus, shouldn't the first fraction be $$\frac{\pu{26.66 ^\circ C}}{50ml}$$ The final answer happens to be $$\pu{26.66 ^\circ C}$$ (which is pretty and convenient), but I think that's just because we chose 5 grams to 10 grams which is a nice number to double and the volume decreases by 5, so everything cancels out nicely.

Kinda of a big segue and separate question but related: What's wrong with the following logic to solve the problem?

$$\pu{50 g}\times \pu{4.184 J g-1 ^\circ C-1} \times \pu{13.33 ^\circ C}=\pu{2789 J}$$

Basically, using q=mcat (m is mass of water, 45, plus mass of solution, 5) to find q of 5 grams $$\ce{CaCl2}$$ and 45 ml water, and then, assuming if we double the mass we double the q, work backwards? (If my logic is wrong, I think it occurs in this assumption)

thus, since we need 10 grams and 40 ml in the end, the mass is still 50 grams, but the 2789 is multiplied by 2 (since mass is doubled) and 13.33C becomes an X.

$$\pu{50 g} \times \pu{4.184 J g-1 C-1} \times x = \pu{5578 J}\\ x = \pu{26.66 ^\circ C}$$

Basically, I think the answer is $$\pu{26.66 ^\circ C}$$ and my teacher thinks it's $$\pu{29.66 ^\circ C}$$.

I'm honestly very confused about q... Google does not seem to help answer this question, if amount of water in a solution/reaction (ie water and $$\ce{CaCl2}$$) changes, but mass of reactant stays the same, does q change?

• Welcome to Chemistry.SE! Please note that formulas can be better expressed with \$\ce{ }\$ for chemical formulas/equations, \$\$ for math term/equations, and \$\pu{ }\$ for units. More information is available in this meta post Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies.
– A.K.
Oct 18, 2018 at 3:38
• Yes, the second ratio is sketchy. The first one is also not quite right except to a first approximation where you say the heat capacity is independent of composition. Nov 11, 2020 at 8:33

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 & =m\cdot C\cdot \Delta T \\ & = 50\cdot4.184\cdot13.33 \\ & = \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 &= \frac{q}{m\cdot C} \\ & \\ &=\frac{5577}{50\cdot4.184} \\ & \\ &= 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

• Thanks for the more mathy way to justify my ideas! Just to clarify, you agree with my proposed change to the ratio and not my teacher's original ratio? Oct 18, 2018 at 11:52
• What makes you think that the heat capacity of a 20% CaCL2 solution is the same as that of water? Oct 18, 2018 at 19:10
• Specific heats of calcium chloride solutions as function of temperature and composition. oxy.com/OurBusinesses/Chemicals/Products/Documents/… Oct 18, 2018 at 19:16
• @ChesterMiller The heat capacity is certainly not the same, I agree, but for a high school AP Chem class, the heat capacity difference is somewhat advanced. Oct 18, 2018 at 22:18
• You first calculate the mass fraction of CaCl2. The heat capacity you use is the average between the values at the initial and final temperatures (determined from the graph in the document). You may need to do this by trial and error, since you don't start out knowing the final temperature. Oct 19, 2018 at 1:00