What is the unknown metal? (specific heat problems)

[Ritzycat]

Homework Statement

Determine the specific heat of the unknown metal in the trial.

This is for a lab, when we put a 167.10 g piece of an unknown metal into a Styrofoam cup that had 107.93g of water in it. Recorded the temperature of the water every 30 seconds until it was the same for two trials in a row. The original temperature of the water was 23°C. (The purpose was to find what element the metal is)

Here is the table of our data. We did two other trials but in both something went horribly wrong.

seconds - °C
0 - 23.0
30 - 23.9
60 - 25.4
90 - 25.8
120 - 25.8

Homework Equations

Q=MCΔT

The Attempt at a Solution

ΔT = 25.8°C - 23.0°C = 2.80°C

This equation is finding the specific heat of water... I think...
Q = (107.93 g)(4.184 J/g°C)(2.80°C) = 1260 J

Now, to find the unknown metal... (our teacher says the possibilities are lead, copper, iron, aluminum, zinc)

1260 J = (167.10g)(x J/g°C)(2.80°C)

1260 J / (167.10g)(2.80°C) = x J/g°C

x J/g°C = 2.69 J/g°C

If I did everything right... 2.69 J/g°C SHOULD be the specific heat value of the unknown metal (since the energy lost by the water is the energy gained by the metal, and vice versa). But none of the values on the PT for heat capacity are 2.69J/g°C.

Did I do something wrong in my calculations, am I interpreting my result wrong, or is there something really bad with my data? I'm starting to think its the latter...

This might help...
http://www2.ucdsb.on.ca/tiss/stretton/Database/Specific_Heat_Capacity_Table.html

All answers & help are appreciated.. This is due tomorrow and I'm completely stumped.

 

[Yanick]

What was the initial temperature of the metal you used? You used the same delta T in both expressions, this may have been your oversight which led to such a discrepancy.

 

[Ritzycat]

Darn it! Yes, I think you got me. We did this lab a few days ago so I was having some trouble recalling the figures of the lab.

The metal was put in water, and the water was heated up until it started to boil. Then we moved the metal into the styrofoam cup and the temperature of the water was recorded every 30 seconds...
Our teacher told us to assume the temperature of the metal would be 100°C in this case since that is when the water boils...

What would be my Delta T in this situation then? The initial temp of the metal (100) and then the final temperature after it has stabilized in the water? Assuming this is so, the Delta T would be 74.2°C and it'd be

1260 J = (167.10g)(x)(74.2°C)

x = .102 J/g°C

A more viable answer but it doesn't really match up with the values for any of them. Closest is gold which is about .27 off. Is there anything else I have to do?

 

[Yanick]

Yes, the final temperature is the temperature at which the system stabilized.

Everything else looks more or less okay to me but I didn't do the calculations myself so you may have an arithmetic error somewhere. Otherwise it looks like your data is just not great. In the future you should re-do trials where things have gone "horribly wrong", for this very reason. Replicates would have helped you make a decision. At this point you may want to just include a good discussion about error and such, showing that you can think about experiments and sources of error is infinitely more important (IMO) than just getting the answer.

 

[Ritzycat]

Thanks for the response. I agree, it probably was something with the error in trial itself, I'll have to recall exactly what it was all about...