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Chemistry 22 Online
OpenStudy (anonymous):

A researcher studying the nutritional value of a new candy places a 5.00-gram sample of the candy inside a bomb calorimeter and combusts it in excess oxygen. The observed temperature increase is 2.25 °C. If the heat capacity of the calorimeter is 39.90 kJ·K–1, how many nutritional Calories are there per gram of the candy?

OpenStudy (goformit100):

@maheshmeghwal9

OpenStudy (anonymous):

\[q=(C)(\Delta T)\] i don't believe the mass has anything to do with this problem because you have the heat capacity for the calorimeter, and by the fact there is nothing to cancell out the grams. Change you degrees C into K

OpenStudy (anonymous):

q= (39.90kj)(275.4)

OpenStudy (anonymous):

That is how i would answer it

OpenStudy (anonymous):

and what to do after i got "q" ?

OpenStudy (anonymous):

you need to find a conversion factor that brings you from kj to cal

OpenStudy (anonymous):

ok so...my answer right now should be in kj, and i have to convert that to J first...then convert from J to Cal correct?

OpenStudy (anonymous):

nvm i just found out that 1 kj = .2390 cal

OpenStudy (anonymous):

did they give you the conversion factors to do that?

OpenStudy (anonymous):

they didn't

OpenStudy (anonymous):

2626 Cal <-- sounds right ?

OpenStudy (anonymous):

(10988.2)(.2390) = 2626

OpenStudy (anonymous):

yea looks right to me however if they have different conversion factors or even a different periodic table our answers are going to differ slightly

OpenStudy (anonymous):

you mean this? 1 Cal = 1000 cal = 4.184 kJ

OpenStudy (anonymous):

yes that

OpenStudy (anonymous):

uh oh...

OpenStudy (anonymous):

how many kj did you get?

OpenStudy (anonymous):

i got 10988 kj

OpenStudy (anonymous):

\[10988 kj (\frac{1 Cal}{4.184kj})=\]

OpenStudy (anonymous):

that still get me 2626.30

OpenStudy (anonymous):

so it's the samething as multiply by .2930 then... :D

OpenStudy (anonymous):

just trust me and always use the conversion factors that are given. Sometimes you will get slightly off looking up other conversion factors because the number could be more or less exact. the periodic table can behave in the same way and because you are doing this on a computer program it's not going to understand why you are coming up with a different answer than it is

OpenStudy (anonymous):

know what i mean?

OpenStudy (anonymous):

yeah i'll keep that in mind thanks :D

OpenStudy (anonymous):

hmmm its wrong ^^!

OpenStudy (anonymous):

maybe that 5 grams have something to do with it

OpenStudy (anonymous):

" how many nutritional Calories are there per gram of the candy?" 2626/5 = 525.2 maybe ?

OpenStudy (anonymous):

yea try that

OpenStudy (anonymous):

omg still wrong lol...wtf I'm gonna go crazy cuz of this x_X

OpenStudy (anonymous):

I would write all the chemistry down and take it to the teacher tomorrow because i'm pretty sure we have been right

OpenStudy (anonymous):

but check your sig figs

OpenStudy (anonymous):

tomorrow is saturday lol :D

OpenStudy (anonymous):

good point, do you have another shot at this question? if so put in 525.500 Cal

OpenStudy (anonymous):

hmm still wrong :(

OpenStudy (anonymous):

it give some explanation: First, use the temperature change and the heat capacity of the calorimeter to calculate the amount of heat produced in kilojoules. Then convert to Calories. Finally, divide your answer by the mass of candy burned to express the energy in Calories per gram.

OpenStudy (anonymous):

we did that

OpenStudy (anonymous):

divide by mass of candy burned ? where is that coming from?

OpenStudy (anonymous):

wth =.=

OpenStudy (anonymous):

try 525.20 Cal

OpenStudy (anonymous):

unless you are out of chances

OpenStudy (anonymous):

i don't think sig fig is necessary in this case

OpenStudy (anonymous):

cuz the other problem i didn't do sig fig either and i got them right...this problem just weird..

OpenStudy (anonymous):

well we did everything the hints told us to do

OpenStudy (anonymous):

ya..wtf.. if i miss a sig fig, it would tell me to fix it..but it didn't say anything about sig fig so i guess that's not it :D

OpenStudy (anonymous):

oh well..i gtg now, thanks for the help zbay cya later!

OpenStudy (anonymous):

sorry i couldn't be more help

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