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Chapter 8: Enthalpy and Thermochemistry

Re: Chapter 8: Enthalpy and Thermochemistry

Postby goldstanda3269 » Fri Mar 11, 2016 3:10 pm

Sheri wrote:Hi GS,

Could you explain q12?
I don't get the explanation there. "Thus, when competing with shapes of equal volume, it will have a smaller surface area than any of the rest. Minimizing the area translates into reducing the size of the numerator while, by hypothesis, the denominator stays the same. This should produce a decrease in Q – an increase in energy efficiency."
which equation is the question/answer referring to when stating numerator and denominator?
Thanks!


The equation being referred to is in the explanation to the previous question. Basically, the first paragraph of the passage was translated into the following equation (yes, even on the real exam, sometimes you will need to produce an equation based on your understanding of the problem):

"From the opening paragraph in the passage, the thermal resistance R = x/(A•k) where x is thickness, A is area and k is thermal conductivity. "

Now thinking in reverse, if we want to have the highest thermal resistance (so this would prevent the loss of heat making the structure most energy efficient), then we want the denominator to be very low (given that there is no change in x and k). A sphere is the smallest surface area relative to volume (this is why droplets of water form spheres; see PHY 6.1.5).

The explanation to this question was a bit confusing, so it has been improved based on the information above.
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby fujiaen » Sat Mar 12, 2016 12:21 pm

Hello,

As I understand it, the equation to measure a change in heat absorbed in a system is given by Q=mc(change in temp).
Why is m (the mass) not included in calculations for Q2?
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby goldstanda3269 » Sat Mar 12, 2016 2:55 pm

fujiaen wrote:Hello,

As I understand it, the equation to measure a change in heat absorbed in a system is given by Q=mc(change in temp).
Why is m (the mass) not included in calculations for Q2?


Here is the simple answer which will help you with dozens of GAMSAT questions: Dimensional analysis. It is not assumed that you would have memorised Q = mcΔT, but it is assumed that you evaluated the units of the heat capacity given in paragraph 2 and, given the temperature, that you realised that you must multiply in order to get kJ which is found among the answer choices. So even if something seems unusual, it must be correct because you followed the units in a disciplined way.

Now going deeper, so this is beyond what ACER expects: Heat capacity (= thermal capacity) is the ratio of the heat added to (or removed from) an object over the resulting temperature change. You were thinking about the "specific heat capacity", often called "specific heat", which is the heat capacity per unit mass of a material. Because ACER will clearly provide the units, these are not details which would be helpful to commit to memory. But keep in mind: they will use a mix of different units and often some conversions will be needed (i.e. kilo/deci/pico/mega/milli/etc.).
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby fujiaen » Sat Mar 12, 2016 7:46 pm

Thanks for the prompt reply!
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby ElleRT » Mon Jun 20, 2016 9:45 pm

I really don't understand this question, how are we meant to know how to do this? Why is force and distance involved?

Which of the following would be consistent with the dimensional formula for the latent heat?

A) MLT -2
B) ML2T -2
C) ML2T -1
D) L2T -2

Result: Your answer is wrong.
Your choice: A Correct choice: D

Explanation: This question is a crossover between general chemistry and physics (which occasionally happens on the real exam).

Q = mL = mLatent (From this point, we will not use L for latent heat so as to avoid confusion with L for length; thus Latent will represent latent heat)

Energy, for example, has the same units as work (i.e. joules) which can be given as (force) x (distance)
(force)(distance) = (mass)(acceleration)(distance) = m Latent = (mass) Latent

Divide both sides by mass:
(acceleration)(distance) = Latent
(m/s2)(m) = m2/s2 = Latent

Replace L for m and T for s:
L2/T2 = L2T -2 = Latent
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby goldstanda3269 » Tue Jun 21, 2016 12:42 am

ElleRT wrote:I really don't understand this question, how are we meant to know how to do this?


You are meant to be able to answer questions like this because of experience.

Every ACER exam has questions based on dimensional analysis (example: GAMSAT Practice Test, Green booklet, Questions 99-101). We have such questions throughout our Physics and General Chemistry chapter review practice questions (also, see Question 10 in our GS GAMSAT Math Chapter 2; in fact, GS GAMSAT Math should be reviewed before studying Physics and Chemistry), as well as our practice exams. Each step is the level of reasoning that is likely to be required on the real exam.

ACER expects that you are fluent with regards to basic SI units.


ElleRT wrote: Why is force and distance involved?


There are many different ways to solve a science problem and we are just presenting what we think is the easiest solution. You are not expected to have memorised specific units for Q (heat energy) but since it is a form of energy, you are definitely expected to know different ways to represent energy, and work, which is force x distance, is one of those ways. Energy and work are reviewed in GS GAMSAT Physics Chapter 5.
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby shaymaa » Thu Nov 03, 2016 7:30 pm

Hello I have question in regards to the following question please:

8) Given the data in Table 1, how much would it cost to maintain the pyramid’s current internal temperature for a full year if no modifications to it were made? (Note: Assume that the pyramid will need to be cooled year round, and that 100,00 btus can be purchased for $0.50)

A) $3,416
B) $159
C) $3,832
D) $1,708
my answer is wrong , so I went through the explanation , could you please explain how did you get the total amount of heat transfer, may be calculated at 2.1 x 106 btu? where did this value come from? and could you please refer which part of the chapter is related to this question, sorry as I couldn't figure it out -:(

many thanks
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby goldstanda3269 » Fri Nov 04, 2016 12:03 pm

RE: Question 8

shaymaa wrote:Hello I have question in regards to the following question please:

. . .

could you please explain how did you get the total amount of heat transfer, may be calculated at 2.1 x 106 btu? where did this value come from?


The numbers in Table 1 are defined in the last paragraph of the passage.

Qb is the total heat transfer through the base in one hour = 2.3 x 10^4 btu
but we want the total amount in 24 hours so: 2.3 x 10^4 btu/hour x 24 hours = 55.2 x 10^4 btu

Qi is the total loss through one side in one hour = 1.6 x 10^4 btu
but we want the total amount in 24 hours AND Figure 1 shows there are 4 sides to a pyramid so:
4 x 1.6 x 10^4 btu/hour x 24 hours = 6.4 x 10^4 btu/hour x 24 hours = 154 x 10^4 btu

154 x 10^4 btu
+ 55.2 x 10^4 btu
----------------------
209 x 10^4 btu
= 2.1 x 10^6 btu


shaymaa wrote: where did this value come from? and could you please refer which part of the chapter is related to this question, sorry as I couldn't figure it out -:(


Most real GAMSAT questions are not based on something specific that you can read in any chapter of any book. Knowledge only gets you used to "hearing the language of science". Most real GAMSAT questions are based on reasoning. You can only do these types of questions correctly by completing many practice questions that require dimensional analysis (a very careful analysis of units) and reasoning. There are no facts that can be memorised to get this question correct.
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby shaymaa » Fri Nov 04, 2016 10:25 pm

thanks a lot am really appreciated :D
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby gamsat773618 » Thu Dec 08, 2016 10:41 pm

Hello,
my q is from the flash cards, general chemistry :

consider (endothermic ): N2o(g)_2NO2(g). using le chetalier's principle. what occurs if there is an increase in temperature?

my understanding is that in endothermic rxn the products' energy is higher than the reactants' energy. if the temperature is increased then so the stress will be on the right side of the reaction(reactants), thus, the direction of the rxn= shift to the left? :( :roll: :? could you please correct me if I was wrong?

Many thanks,
Shaymaa
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby goldstanda3269 » Fri Dec 09, 2016 7:06 pm

gamsat773618 wrote:Hello,
my q is from the flash cards, general chemistry :

consider (endothermic ): N2o(g)_2NO2(g). using le chetalier's principle. what occurs if there is an increase in temperature?

my understanding is that in endothermic rxn the products' energy is higher than the reactants' energy. if the temperature is increased then so the stress will be on the right side of the reaction(reactants), thus, the direction of the rxn= shift to the left? :( :roll: :? could you please correct me if I was wrong?


Endothermic Reaction: energy is added to the reactant so that the reaction can proceed to make a product:

Energy + reactant <----> product

By conservation of energy: Total energy must be conserved which means that the product's energy is indeed higher than the reactant's energy. An increase in temperature is the same as increasing energy which means that the stress on the LEFT side which makes a shift to the right producing more product (which is consistent with the answer from the flashcard).

It is useful for these types of products to add energy to either the reactant side (endothermic) or product side (exothermic) so that it is easy to visualise what happens with a temperature/energy-related stress to the reaction.
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby gamsat773618 » Fri Dec 09, 2016 11:37 pm

thanks a lot for the exaplanation :D

regards
shaymaa
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby horbowiec.7527 » Sun Jun 27, 2021 12:52 pm

This is regarding GEN CHM Chapter 7 Worked solution for Q22 (CHM-201):
Worked solution state answer A is correct but in explanation it says "ΔG is now positive; ΔH is now negative; ΔS is now negative." so assuming answer D is correct not A?

Regards,
Ania
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby horbowiec.7527 » Sun Jun 27, 2021 1:02 pm

This is still regarding GEN CHM Chapter 7 Worked solution for Q22 (CHM-201):

Why there is an assumption in worked solutions that if the reaction "occurs quickly in room temperature" then it is spontaneous?

I thought you can have spontaneous reactions which occur more slowly but they still favor product formation so will be spontaneous?

Regards
Ania
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Re: Chapter 8: Enthalpy and Thermochemistry

Postby goldstanda3269 » Sun Jun 27, 2021 6:22 pm

horbowiec.7527 wrote:This is regarding GEN CHM Chapter 7 Worked solution for Q22 (CHM-201):
Worked solution state answer A is correct but in explanation it says "ΔG is now positive; ΔH is now negative; ΔS is now negative." so assuming answer D is correct not A?


Yes, that is correct and it has been updated.
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