Specific heat capacity of brass

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SUMMARY

The specific heat capacity of brass was calculated using a copper calorimeter, with the experiment yielding a result of 3.8174 J/g°C, which is incorrect compared to the expected value of 0.3817 J/g°C. The calculations involved the mass of the brass bob (32.5 g), calorimeter (39.7 g), and water (54.2 g), along with their respective specific heats. The discrepancy in results suggests potential errors in data collection or experimental setup, particularly regarding temperature measurements and mass accuracy.

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emmfranklin
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Homework Statement


i getting the wrong the answer

i am

trying to find the specific heat capacity of brass
using copper calorimeter
Data :
mass of brass bob= 32.5gm
mass of calorimeter = 39.7 gm
mass of water + calorimeter = 93.9gm

mass of water = 93.9 - 39.7 = 54.2g

specific heat of water = 4.2 j/gm deg Cel
specific heat of copper = 0.382 j/gm deg cel

Temp of water = 23 deg cel
Temp of brass = 94 deg cel

temp of mixture = 47 deg cel




Homework Equations


formulas used..

part A

(Mc*cc + Mw*cw)(T1-T2)

(mass of the calorimeter* specific heat of the calorimeter + mass of the water*specific heat of the water)*Fall in the temperature.


Part B
C=H/mΔT



The Attempt at a Solution



part A being solved

(mass of the calorimeter* specific heat of the calorimeter + mass of the water*specific heat of the water)*Fall in the temperature.

(39.7 * 0.386) + (54.2 * 4.2) * (47-23)
(15.3242 + 227.64) * 24

242.9642 * 24

5831.1408

part b being solved.

C=H/mΔT
= 5831.1408/ 32.5 * 47

= 5831.1408 / 1527.5

= 3.8174407856

specific heat of brass = 3.8174407856 j/gm deg cel




this answer is wrong

the expected answer is 0.3817

where am i going wrong...
 
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The SHC of brass should be roughly similar to that of copper because brass is made from copper so the book answer is probably correct and google confirms it..

In which case the heat lost by the brass is

E = 0.3817 * 32.5 * (94-47) = 583 Joules

If the water (alone) rises from 23 to 47 then the energy gained by the water is

E = 4.2 * 54.2 * (47-23) = 5463 Joules

so I think there is something wrong with the original data.
 
hi CWatters

I and my colleague tried this experiment many times . but we come to almost the same wrong answer.

we are not able to find where could the data go wrong.
 
The result would come out better if the mass of the brass happened to be 10x higher (325gm rather than 32.5gm). Possible transcription error?
 
Thanks sir for your prompt reply

But I checked the mass there is no transcription error

It is 32.5 grams

But similarly I too doubt that somewhere there is an issue with a decimal point

But I can't figure out
 
emmfranklin said:
Thanks sir for your prompt reply

But I checked the mass there is no transcription error

It is 32.5 grams

But similarly I too doubt that somewhere there is an issue with a decimal point

But I can't figure out

Is this a problem from a text, or is it data reduction from a laboratory experiment that you have performed? You mention that you and your colleague "tried the experiment many times"...
 
It is a data from a lab experiment
 
emmfranklin said:
It is a data from a lab experiment

Ah. I see. After examining the procedure and the equipment and after a trial run, did you make a list of possible sources of error? Is there any place that heat might come from or escape to during the experimental runs?
 
Is it possible the thermocouple is measuring the temperature of the water close to the brass bob and that it's not 47C everywhere?
 
  • #10
Thanks again

I'll verify the points you suggested and get back to you later

And also go through possible errors
 

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