Chemistry Asked by catdog on December 7, 2020

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 segway 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?**

The ratio strategy will definitely work, but I think going back to $q=mcdot Ccdot Delta T$ 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 $Delta T = 13.33$.

$$begin{align} q & =mcdot Ccdot Delta T \ & = 50cdot4.184cdot13.33 \ & = ce{2789 J} end{align}$$

**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 $ce{Delta T = frac{q}{mcdot C}}$. We know $q = 5577$, $m=50$, and $C=4.184$.

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

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

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

$q_1 / T_1 = mcdot C$ and $q_2 / T_2 = mcdot C$.

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

Answered by Christopher Marley on December 7, 2020

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