Question

help with enthalpy

Answer 1

What exactly?

Answer 2

i have no idea what this question is even asking for so:
Part I:
1. Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

qwater = m × c × ΔT.

We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL.

The water has absorbed the heat of the metal. So, qwater = qmetal
2. Using the formula qmetal = m × c × ΔT, calculate the specific heat of the metal. Use the data from your experiment for the metal in your calculation.

Answer 3

perfect... thank you open study update </3

Answer 4

You have to set up a q and negative q equation

Answer 5

So -q =q

Answer 6

Q=mCdelta t

Answer 7

Exactly, now input the values

Answer 8

Remember temperature is in this case, final minus initial

Answer 9

Yeah that one .-. idk what theyre asking for
known metal
Measured mass of metal
27.776
Distilled water volume
25
Distilled water temperature
25.5
Temperature of metal
100.5
Temperature of mixture
38.9

Answer 10

what do you ean final?

Answer 11

Final is the temperature after the occurrence of the experiment and initial is the start temp

Answer 12

yay your a yellow now <3

Answer 13

so the initial temp is 25.5 and im guessing the mixtures temp is the final?

Answer 14

Oh thanks haha

Answer 15

your welcome <3

Answer 16

Seems to be in this case if i stand corrected

Answer 17

well if it isnt the mixture than there is no final temperature so... well go with the mixture ^.^

Answer 18

Yea haha

Answer 19

so its just 38.9-25.5? thats easy 13.4

Answer 20

what i dont get is that whole equation. what is it asking for?

Answer 21

Yea but then u need to calculate the wwters energy in joules and it has to be set equal to the mixtures

Answer 22

so it would be 3.18J/(g x 13.4)?

Answer 23

For the mixture or water?

Answer 24

o.o i thought there was just one thing. thats what the 13.4 was for. like final - initial

Answer 25

Where did u get. 3.18 from?

Answer 26

*4.18 sorry

Answer 27

Ok yea it is

Answer 28

so there is only one thing to solve... and besides the 4.18 my equation is correct?

Answer 29

Wait but didnt it say that you are trying to find the specific heat of the metal?

Answer 30

erm.... i think its water "Calculate the energy change (q) of the surroundings (water) using the enthalpy equation"
not too positive though

Answer 31

heh

Answer 32

http://prntscr.com/4mv4ok

Answer 33

Hey aaron prison buddy! Remember?

Answer 34

so for the first part,

\(q_{water}=m*C_p*[T_f-T_i]=25~g*(4.18~J/g*C^o)*[38.9-25.5]C^o\)

is this what you have?

Answer 35

haha hi arly

Answer 36

Who is that? Thats not my name !

Answer 37

i mean, hi stranger

Answer 38

I think its the wrong person you are thinking of

Answer 39

i think so... haha sorry.

So is 2. regarding Part I or Part II, @lovelyharmonics ?

Answer 40

part one

Answer 41

oh you changed your profile picture i didnt recognize you

Answer 42

haha i did. okay, so you already established that the heat absorbed by the water was released by the metal which is expressed mathematically with:

\(-q_{metal}=q_{water}\)

right?

Answer 43

woah. thanks to studious i now realize that your name is aaron and your last name starts with a q....

Answer 44

and why thats so creepy is because we share the same name o.o different spelling but the same name none the less

Answer 45

haha you have the girl version, with an "e"?

Answer 46

haha OS revelations today

Answer 47

yeah erin <3 were related by name its official

Answer 48

Flirting i see?

Answer 49

Oops thats to aaronq

Answer 50

woo same name party!

Answer 51

no. i woudnt ever date someone with the same name as me. itd be too confusing when someone said "hey erin!" wed both turn to see whos talking and itd be a mess

Answer 52

that would be really confusing hah back to the question!

Answer 53

Lonely im ms lonely

Answer 54

yeah i tots did that though

Answer 55

haha be quiet you birdo

Answer 56

Third wheel haha

Answer 57

shes cool. shes with me

Answer 58

Haha be quiet you birdo

Answer 59

haha im just joking.

Answer 60

Parrot to specific

Answer 61

To be haha

Answer 62

dan is on your question you is good. you is fine.

Answer 63

you wanna use the quantity of heat you got from the first part.

\(-q_{water}=27.776~g*C_p*[38.9-100.5]C^o\)

Answer 64

holy crap where did that just come from

Answer 65

i type fast, what can i say

Answer 66

no i mean where did the numbers just come from

Answer 67

from the file you posted.. :P

Answer 68

^there are several people already working on it, i'd hate to just intervene. If it's not resolved after they're through i will take a look.

Answer 69

i hate it when people inturrupt my question like thats what pm is for.... hey wait where did studious go .-.

Answer 70

i know, its pretty rude. um idk, her ipad crashes all the time, i'm sure she'll be back

Answer 71

shes back.. okay but seriously like you just conjured those numbers out of thin air....

Answer 72

haha i just made them up... lol no really, look at the file you posted.

The mass of the metal is there, and the initial temp as well as the final temp (as the temp of the mixture).

Answer 73

Hey dont cry you guys im back, and you aaron why are you talking about me behind my back, rude jk

Answer 74

wait what is C_p

Answer 75

specific heat capacity

Answer 76

sorry, i couldn't resist talking smack while you were gone.. it's just too easy

Answer 77

well thats 4.18J/(gxC)

Answer 78

that's for the water

Answer 79

i thought we were calculating the water

Answer 80

Haha yea bc u have a big mouth oooooooooo do you need some ice for that burn?

Answer 81

nope, that was 1.

in 2. we're using the data of the metal to find it's specific heat capacity

Answer 82

I PICKED ONE OF THESE WHICH ONE IT IS IDK http://prntscr.com/4mvgig

Answer 83

i need some liquid nitrogen for that burn, please

Answer 84

Hahaeven better..... DREADS hahah

Answer 85

kids, play nice now

Answer 86

Inside joke

Answer 87

Haha ok mommy but hes being mean

Answer 88

twin, quit being mean