Homework 8.25

Moderators: Chem_Mod, Chem_Admin

Natalie Yakobian
Posts: 45
Joined: Fri Sep 25, 2015 3:00 am

Homework 8.25

Postby Natalie Yakobian » Sat Jan 09, 2016 3:38 pm

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
Posts: 13
Joined: Fri Sep 25, 2015 3:00 am

Re: Homework 8.25

Postby Tara Kubilay 2A » Sat Jan 09, 2016 5:25 pm

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
Posts: 3
Joined: Fri Jun 17, 2016 11:28 am

Re: Homework 8.25

Postby Chris Prom » Thu Jan 19, 2017 4:23 pm

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
Posts: 30
Joined: Wed Sep 21, 2016 2:58 pm

Re: Homework 8.25

Postby NinaSheridan » Fri Jan 20, 2017 12:30 am

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
Posts: 19
Joined: Wed Sep 21, 2016 2:56 pm

Re: Homework 8.25

Postby Hue_Vo_1D » Sat Jan 21, 2017 11:49 pm

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
Posts: 6
Joined: Wed Sep 21, 2016 2:57 pm

Re: Homework 8.25

Postby genesis_escobar_2M » Sat Jan 28, 2017 12:38 pm

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


Return to “Heat Capacities, Calorimeters & Calorimetry Calculations”

Who is online

Users browsing this forum: No registered users and 1 guest