Homework 8.25

[Natalie Yakobian]

Hi,
I was wondering how to solve homework question 8.25.

"A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat
in 0.200 L of solution in the calorimeter (q= -3.50 kJ), resulting in a temperature rise of 7.32 C. In a subsequent experiment, 100.0 mL of 0.200 m HBr(aq) and 100.0 mL of 0.200 m KOH(aq) were mixed in the same calorimeter and the temperature rose by 2.49 C. What is the change in the internal energy of the neutralization reaction?"

When I multiplied the proportion of temperature change to q (-3.50 / 7.32) by the delta T of the subsequent reaction (2.49) I got the correct answer but don't really understand why. Is this even the correct way to solve the problem?

Thank you so much.

[Tara Kubilay 2A]

What you are doing is correct. The proportion of temperature change to the released energy (-3.50kJ/7.32 C) gives you the heat capacity of the calorimeter, because heat capacity is " the amount of energy required to raise the temperature of an object one degree." And it is a constant for the same object at the same temperature and pressure. You are actually using the equation C=q/(T(final)-T(initial)). Second experiment uses the same calorimeter, so you use the same heat capacity and the same equation with the new temperature change to find the the change in the internal energy of the neutralization reaction (q).

[Chris Prom]

Follow up question:
How do we know that we only need the Heat Capacity,
and not specific heat capacity for all substances at work (Hbr, KOH, and 1 unknown substance)

[NinaSheridan]

Also why do we completely ignore the masses that are given? Isn't the equation (m)(c)(deltaT)? Why do we disregard the masses and the moles?

[Hue_Vo_1D]

From what we are given, the reaction in the calorimeter releases 3.50 kJ of heat, which resulting in a temperature rise of 7.32 C in the calorimeter. The first step in solving this is -3.50kJ/7.32 C, which gives us .478 kJ/C; this is the heat capacity of this one specific calorimeter that is used in these experiments. It takes .478 kJ to rise the temperature by 1 C in the calorimeter. Like Tara said above, "the second experiment uses the same calorimeter, so you use the same heat capacity."

After this first step, we can reword the question as if .478 kJ is required to raise the temperature of the calorimeter by 1 C, how much heat is released if there's the rise in temperature by 2.49 C in the same calorimeter? (multiplying)

Note: heat capacity is the amount of energy required to raise the temperature of an object(in this case the calorimeter) one degree.

What we're looking at in this question is more about heat capacity of the calorimeter itself rather than specific heat capacity or molar heat capacity of single substances in the reaction.

Like Dr.Lavelle just answered in one of the recent questions,
"For calorimeters, we are only interested in its heat capacity."

[genesis_escobar_2M]

do we just avoid the negative sign when calculating heat capacity?

[Ivan Huang Dis 3B]

if the temperature rises why is the change in energy negative not positive. Wouldn't positive indicate an endothermic reaction where it absorbed heat?

[Isabelle Kim 3E]

if the temperature rises why is the change in energy negative not positive. Wouldn't positive indicate an endothermic reaction where it absorbed heat?

I may be incorrect, but isn’t it due to the fact that the final temperature being higher indicates that heat energy has been released during the chemical reaction. Therefore, the overall change in energy would be negative when we view the heat exchange between the system and surroundings. Or in other words, we are viewing the heat energy being released from the system into its direct surroundings and that is why there is a negative sign. Somebody correct me if I got it all wrong!

[Brooklyn Burgess 3L]

if the temperature rises why is the change in energy negative not positive. Wouldn't positive indicate an endothermic reaction where it absorbed heat?

The temperature in the calorimeter increases, so the reaction is releasing heat. Therefore, q of the reaction is negative.

[Brooklyn Burgess 3L]

Why do we ignore the given liters/milliliters when solving this question?

[Ashwin Vasudevan 3A]

Do we just ignore 100.0 mL of 0.200 M HBr(aq) and 100.0 mL of 0.200M KOH or is that important to answering the question?

[alexjung1A]

Do we just ignore 100.0 mL of 0.200 M HBr(aq) and 100.0 mL of 0.200M KOH or is that important to answering the question?

I believe we can just ignore those measurements.

[sophiavmr]

I am confused about what q=-3.50 is representing. I believe that it is the heat released by the reaction, in which case q=-q(cal). Therefore, the heat energy of the calorimeter would be positive 3.50

[WS405590915]

I am confused about what q=-3.50 is representing. I believe that it is the heat released by the reaction, in which case q=-q(cal). Therefore, the heat energy of the calorimeter would be positive 3.50

from my understanding q=-3.50 is the heat released from the reaction inside the Calorimeter. You would be correct to assume that the (delta)q for the Calorimeter is positive 3.50, since the heat has to go somewhere. However for the context of the question, when talking about internal energy, we aren't as concerned with the Calorimeter as much as we are with what happens inside. Thus the q value to use to solve should be -3.50.

[Deivy Gonzalez 3H]

I was not sure why they gave us those measurements such as 100 mL. I was trying to incorporate them in some type of equations. But thank you for saying to just ignore them.

[Jayla Riley]

We can ignore the other unnecessary variables that do not effectively contribute to the answer of the change in internal energy because the reaction does no work as all energy is conserved within the system (calorimeter) and it is translated into heat released. Therefore, in this case, q and delta U are the same.

[205984433]

I don't understand this problem at all. Why can we disconsider the mols in q=ncdeltaT?