Question

https://www.apchemsolutions.com/downloads/thermo-2-ws.pdf number one @vocaloid

Answer 1

a) heat gained by the copper: q = mc deltaT
since we're near room temperature we assume water has a density of 1g/1mL
so we have

425g of water * 4.186 J/gC * (26.15-22.55)

Answer 2

I'm guessing, but is a) 94.23 K

Answer 3

Ah wait, lemme work this out, thank you so much

Answer 4

K is a unit of temperature, not heat
using the specific heat capacity i've used (4.186 J/gC) will give joules as the final heat units

Answer 5

Why do we use the water for total mass, not the copper?

Answer 6

Oh wait, I see. A is asking for the change in water. Sorry about that.

Answer 7

for b) for a closed system we assume no energy has been lost, so all the energy gained by the water is lost by the copper

Answer 8

c) I'm not 100% sure but i believe it would just be the same logic as a)

q = mc delta T
(your solution from part a) = 85.2 * C * (221.32 - 22.55)

Answer 9

Wait so A and B are both 6404.6 J if I calculated correctly?

Answer 10

Or the temperature change is the same for both A and B

Answer 11

heat change is the same
temp change I believe is also the same (I'm assuming the system is at thermal equilibrium)

Answer 12

I don't know if I got this right, I'll put in the work so it's easier for you

Answer 13

I don't know if I got this right, I'll put in the work so it's easier for you

Answer 14

Part A
q=mc∆t
q=(425g)(4.186J/gC)(26.15C-22.15C)
q=6404.58 J

Part B
q=mc∆t
q=(85.2g)(0.385J/gC)(26.15C-221.32C)
q=-6401.96J

Answer 15

yeah the calculations are good

Answer 16

they might have wanted you to calculate the specific heat of copper rather than looking it up, by setting -6404.58 J = 82.5 * C* (26.15C-221.32C) but that's just my reasoning

Answer 17

since that's what part C is asking

Answer 18

Oh really. How would you find a specific heat of an element?

Answer 19

like, if we assume that all the heat gained by the water is lost by the calorimeter we just set

q = mcdeltaT and solve for C this time instead of q

Answer 20

idk i could just be overthinking it

Answer 21

Oh Wait I get it. Since no energy is supposed to be lost in a system because it's a calorimeter and it's not supposed to, the answer of B is supposed to be the opposite of A.

Answer 22

yeah theoretically

Answer 23

And how would you approach D?
I thought of it as kinetic and potential energy but I'm confused between the difference of the two and what it has to do with it.

Answer 24

as a general rule energy is always conserved; the energy lost from the metal is transferred to the water, showing conservation. even if the transfer wasn't perfect, the lost energy would be dissipated into the environment but still ultimately conserved

Answer 25

Oh thank you so much!!!!

Answer 26

are you gonna try the other ones next?

Answer 27

Yeah Idk how to start it at all. Is this the one where you create a table of multiple reactions? I don't know how to do those.

Answer 28

I don't know how to decipher what problems use the bond enthalpy crap, the hess's law table crap, or the ∆H degree crap

Answer 29

alright, for 2a) we just use q = mcdeltaT as before
you have 200mL total of solution, and you are given the density 1g/1mL, so you just have 200g

then it's just 200g * 4.184 K/gK * (28.2-23)

Answer 30

after that, heat of reaction is typically given as kJ/mol, so you'd just convert Joules from the previous step to kJ then divide by the moles of product formed

Answer 31

still there? maybe there's something conceptual I forgot to explain

enthalpy of formation is the heat (usu. given in kJ) released when 1 mole of a compound is formed

the reaction in this problem is the formation of water from acid and base

so you'd calculate total heat using q = mcdeltaT, treating the entire solution as your system, then divide by the moles of H2O formed

since we have 100mL of a 0.76M solution of acid that gives us

0.76 = moles/.1L, or 0.076 moles acid, same thing with base, 0.076 moles base, forming 0.076 moles product
so whatever your q value you get, divide that by 0.076 moles product to get the heat of formation

Answer 32

Voca is inky too omg