Chemistry, 28.06.2020 04:01 Rixi1227

The same reaction, with exactly the same amount of reactant, is conducted in a bomb calorimeter and in a coffee-cup calorimeter. In one measurement, qrxn=−13.3qrxn=−13.3 kJkJ in the other qrxn=−11.9qrxn=−11.9 kJkJ. Which value was obtained in the bomb calorimeter? (Assume that the reaction has a positive ΔVΔV in the coffee-cup calorimeter.)

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Answer from: ellisc7044

$q_{rxn}$ = -13.3 KJ belongs to the bomb calorimeter

Explanation:

In this question, we are concerned with determining which of the two heats of reaction given, belongs to that in the bomb calorimeter

The kind of calorimetry used in a bomb calorimeter and a coffee-cup calorimeter is different.

In a bomb calorimeter, what we have is a constant volume calorimetry while in a coffee-cup calorimeter, what we have is a constant-pressure calorimetry.

For constant volume calorimetry, the change in volume ΔV = 0 while for constant pressure calorimetry, the change in pressure ΔP = 0.

Mathematically, the relationship between the change in internal energy Δ $E_{rxn}$ of the reaction, the work of the reaction $w_{rxn}$ and the heat of the reaction $q_{rxn}$ is given by;

Δ $E_{rxn}$ = $w_{rxn}$ + $q_{rxn}$

Kindly note that, in a bomb calorimeter, the work of the reaction $w_{rxn}$ = 0

Thus, this means that for a bomb calorimeter,

Δ $E_{rxn}$ = $q_{rxn}$

This means that in the bomb calorimeter, the entire internal energy change is converted to heat of the reaction.

In the case of the coffee-cup calorimeter, the work of the reaction is non-zero and it is equal to the product of the pressure and the change in volume.

Thus, mathematically,

$w_{rxn}$ = PΔV

Thus;

Δ $E_{rxn}$ = $w_{rxn}$ + $q_{rxn}$ = PΔV + $q_{rxn}$

Since for a coffee cup calorimeter, the change in volume is positive(as given in the question), this means that;

ΔV > 0

and PΔV > 0

This automatically means that

Δ $E_{rxn}$ > $q_{rxn}$

Hence, in the coffee cup calorimeter, the entire change in internal energy is not converted to heat as part is used to do useful work.

Thus, this means that in the coffee cup calorimeter, the value of change in internal energy is greater than the value of heat.

This makes the magnitude of heat in the bomb calorimeter greater than that of the coffee cup calorimeter.

Since -13.3 KJ is the bigger magnitude of the two, then it belongs to the bomb calorimeter

Answer from: Quest

The answer to your question really depends on three things: how much water you have, how cold it is to start out, and how cold the things around it are. water actually freezes when it gets to 32 degrees fahrenheit (0 degrees celsius), but the time it takes to get there may be different. let's start with the first. if you take two glasses, and fill one with a tiny bit of water, and the other about halfway, then put them both in the freezer, the one with less water will freeze first (you can try this at home, but i recommend using plastic cups and not glass ones). now let's move on to the second part. let's say you have two glasses, and you fill one with really cold water that has been in the refrigerator, and the other with really hot water from the sink. if you put both of them in the freezer, the one that started out colder will freeze first. for the third part, let's imagine that you have two glasses with the same amount of water in them, and the water is at the same temperature. imagine putting one outside on a really really cold day in georgia, and having a friend in alaska put one outside on the same day. since it would be so much colder in alaska, the glass of water there would freeze before yours. so, if you took a tiny bit of really cold water in a glass, and put it outside on a cold day in alaska, it would freeze a lot faster than a big glass of hot water outside on a cold day in georgia.

Answer from: Quest

The correct answer is b) false