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Calorimetry Lab Analysis (predict specific heat of unknown metal)

19,081
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well the formula ive been using doesnt seem to give me the right answer
it gives me 20 something J/gC
when i should be getting less than 1 J/gC
So any other wild ass choice of an equation is fair game?

The reason you are getting the wrong answer is not because of the equation you are implementing. It is because of the wrong way that you are implementing the equation. In each of the cases, please list the initial and final temperatures (all in degrees C) of the cube and of the water. Thanks.
 
So any other wild ass choice of an equation is fair game?

The reason you are getting the wrong answer is not because of the equation you are implementing. It is because of the wrong way that you are implementing the equation. In each of the cases, please list the initial and final temperatures (all in degrees C) of the cube and of the water. Thanks.
umm i did

dont look at the bottom three trials
those are for lead
the teacher told us that was lead so we could test our procedure on that as practice

the first three trials are the ones about the unknown metal cube he gave us
i filled in all the information you need
 
19,081
3,735
the initial temp of the cube if before i put it into the boiling water
the final temp of the cube is after i put it in the boiling water for a couple minutes
the initial temp of the water is room temp (before i put the boiling cube in)
the final temp of the water is the highest temp the room temp water becomes after putting the cube in
According to your data sheet (as I understand it), the water temperature rose to about 100 C, and the cube temperature dropped to about 26 C. Is this a correct interpretation? Or was the final temperature of both the water and the cube 26 C?
 
gosh dang it
i am so dumb
the initial temp for the cube is 100C not 21C
lkajsdfoanoginlkadjfkalkdlknkladjglkalkdjflkajlkdjlkgjaklsdjglajlgd
i wasted so much time of my life repeatedly putting this wrong information into my calculator
 
now my specific heat is negative 1
great
lol
 

ZapperZ

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do you think i need ideal gas law?
I have no idea, because I don't know what you are doing. Besides, this is not relevant to your question in this thread.

and do you think you can help me with the thing on google sheets?
A what?

Zz.
 
19,081
3,735
$$C_{cube}=\frac{(30)(4.184)(26-20)}{(121.5)(100-26)}$$

????
 
I have no idea, because I don't know what you are doing. Besides, this is not relevant to your question in this thread.
well it turns out i dont need it as someone else explained to me

and google sheets is the thing that is kinda like microsoft excel that lets you make spreadsheets
 
now we just need to find it for the other two trials, average it, and then find the metal with that specific heat

thank you so much!!!
 
but i fixed the spreadsheet data and now im getting 1.1 J/gC
calorimetry is so weird
 
4.184*30*76.8/73.3/121.5=1.082
4.184*30*79.3/74/121.5=1.107
4.184*30*75.8/72.8/121.5=1.076

???
 
19,081
3,735
4.184*30*76.8/73.3/121.5=1.082
4.184*30*79.3/74/121.5=1.107
4.184*30*75.8/72.8/121.5=1.076

???
is the 76.8 in degrees C or degrees F? Is the 73.3 in degrees C or degrees F? What are the actual initial and final temperatures of cube and water for this case?
 
is the 76.8 in degrees C or degrees F? Is the 73.3 in degrees C or degrees F? What are the actual initial and final temperatures of cube and water for this case?
the 76.8 and 73.3 are changes in temperature
 
is the 76.8 in degrees C or degrees F? Is the 73.3 in degrees C or degrees F? What are the actual initial and final temperatures of cube and water for this case?
the actual are
trial namemass in gti in Ctf in Cmass water in gti water in Ctf water in Cct in Cct water in Cspecific heat in J/gC
UM1121.599.626.33022.899.673.376.81.082415239
 
acronyms
ti initial temperature
tffinal temperature
ggrams
Cdegrees celcius
ctchange in temperature
Jjoules
UMunknown metal trial
LTlead trial
 
19,081
3,735
How can the final temperature of the water be higher than the final temperature of the cube?
 
the initial temp of the cube if before i put it into the boiling water
the final temp of the cube is after i put it in the boiling water for a couple minutes
the initial temp of the water is room temp (before i put the boiling cube in)
the final temp of the water is the highest temp the room temp water becomes after putting the cube in
 

tech99

Gold Member
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the water ti and tf is how much the room temperature water was heated once the metal (at 212 C) was placed into it. so it goes from about 21 to 26 because the water has high specific heat so it is hard to raise its temperature even if a block of boiling metal is placed into it.
the metal ti and tf however is how much the metal was heated once placed into boiling water. so that would be from 21 to 212.

as i said earlier the procedure went as follows:
put metal into boiling water
metal is now at boiling temperature
put that same metal into room temp water
see how much the room temp water heats
calculate specific heat of metal
use that to identify the metal
You heated the water to 100C not 212C.
 
19,081
3,735
the initial temp of the cube if before i put it into the boiling water
the final temp of the cube is after i put it in the boiling water for a couple minutes
Are you aware that this has nothing to do with the measurement you are trying to make? The only reason for putting the cube in the boiling water was to establish its starting temperature for the subsequent measurement.
 
You heated the water to 100C not 212C.
yea i fixed that already
but i cant change the original post for some reason
i changed my spreadsheet though
 
Are you aware that this has nothing to do with the measurement you are trying to make? The only reason for putting the cube in the boiling water was to establish its starting temperature for the subsequent measurement.
oh
so would the new one be
water:
21 --> 26
metal:
212 --> 26
 

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