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

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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