589.55= (.40005*1.50)/(.05345*79.60)- Our Results
Uncertainty = (0.005/.40005 + 0.10/1.50 + 0.005/.05345 + 0.10/79.60)589.55
Uncertainty = 102.5621
The result is 589.55 ± 100 Hence our range is from 489.55 to 689.55
Trial 2 – Brass
To find the specific heat of Brass we have to reorder mh2och2o(Ti-Te)metal=mmetalcmetal(Ti-Te)metal to:
cmetal= (mh2o*ch2o*(Ti-Te)h2o)/(mmetal*(Ti-Te)metal)
cmetal= (.401*4180*2.20)/(.05348*80.00)
cmetal= (3687.596)/(42.784)
cmetal= 861.91
To find the percentage error for our specific heat we use the following formula:
|Accepted Value – Our Value | / (Accepted Value * 100)
|380 – 861.91| / 380 * 100
581.91/38000
126.82%
The error for Brass on our first trial is 126.82%
To find the uncertainties in our measurement we reorder the following formula Error Q / q = (Error a/a + Error b/b…) to Error Q = (Error a/a + Error b/b…)*q. Because we used cmetal= (mh2o*ch2o*(Ti-Te)h2o)/(mmetal*(Ti-Te)metal) we need to add the estimated error of Te and Ti and use the formula above for the multiplications.
cmetal= (mh2o*(Ti-Te)h2o)/(mmetal*(Ti-Te)metal) – We ignore the specific heat of H2O because no error is present.
cmetal= (0 .005* 0.10)/( 0.005*0.10) – Our Error Values
861.91= (.401*2.20)/(.05348*80.00) - Our Results
Uncertainty = (0.005/.401. + 0.10/2.20 + 0.005/.05348 + 0.10/80.00)861.91
Uncertainty = 131.5846
The result is 861.91± 100 Hence our range is from 761.91to 961.91
Steal Table 2
Trial 1 – Steal
To find the specific heat of Steal we have to reorder mh2och2o(Ti-Te)metal=mmetalcmetal(Ti-Te)metal to:
cmetal= (mh2o*ch2o*(Ti-Te)h2o)/(mmetal*(Ti-Te)metal)
cmetal= (.40005*4180*1.10)/(.04945*80.00)
cmetal= (1839.4299)/(3.956)
cmetal= 464.97
To find the percentage error for our specific heat we use the following formula:
|Accepted Value – Our Value | / (Accepted Value * 100)
|380 – 464.97| / 380 * 100
84.97/38000
22.36%
The error for Steal on our first trial is 22.36%
To find the uncertainties in our measurement we reorder the following formula Error Q / q = (Error a/a + Error b/b…) to Error Q = (Error a/a + Error b/b…)*q. Because we used cmetal= (mh2o*ch2o*(Ti-Te)h2o)/(mmetal*(Ti-Te)metal) we need to add the estimated error of Te and Ti and use the formula above for the multiplications.
cmetal= (mh2o*(Ti-Te)h2o)/(mmetal*(Ti-Te)metal) – We ignore the specific heat of H2O because no error is present.
cmetal= (0 .005* 0.10)/( 0.005*0.10) – Our Error Values
464.97= (.40005*1.10)/(.04945*80.00) - Our Results
Uncertainty = (0.005/.40005 + 0.10/1.10 + 0.005/.04945 + 0.10/80.00)464.97
Uncertainty = 95.67677
The result is 464.97± 100. Hence our range is from 364.97to 564.97
Trial 2 – Steal
To find the specific heat of Steal we have to reorder mh2och2o(Ti-Te)metal=mmetalcmetal(Ti-Te)metal to:
cmetal= (mh2o*ch2o*(Ti-Te)h2o)/(mmetal*(Ti-Te)metal)
cmetal= (.3998*4180*3.00)/(.0495*80.10)
cmetal= (5013.492)/(3.964)
cmetal= 1264.45
To find the percentage error for our specific heat we use the following formula:
|Accepted Value – Our Value | / (Accepted Value * 100)
|380 – 1264.45| / 380 * 100
884.45/38000
232.75%
The error for Steal on our first trial is 232.75%
To find the uncertainties in our measurement we reorder the following formula Error Q / q = (Error a/a + Error b/b…) to Error Q = (Error a/a + Error b/b…)*q. Because we used cmetal= (mh2o*ch2o*(Ti-Te)h2o)/(mmetal*(Ti-Te)metal) we need to add the estimated error of Te and Ti and use the formula above for the multiplications.
cmetal= (mh2o*(Ti-Te)h2o)/(mmetal*(Ti-Te)metal) – We ignore the specific heat of H2O because no error is present.
cmetal= (0 .005* 0.10)/( 0.005*0.10) – Our Error Values
1264.45= (.3998*3.00)/(.0495*80.10) - Our Results
Uncertainty = (0.005/.3998+ 0.10/3.00 + 0.005/.0495 + 0.10/80.10)1264.45
Uncertainty = 187.2627
The result is 1264.45 ± 200. Hence our range is from 1064.45 to 1464.45
Aluminum Table 2
Trial 1 – Aluminum
To find the specific heat of Aluminum we have to reorder mh2och2o(Ti-Te)metal=mmetalcmetal(Ti-Te)metal to:
cmetal= (mh2o*ch2o*(Ti-Te)h2o)/(mmetal*(Ti-Te)metal)
cmetal= (.40005*4180*1.30)/(.01795*79.80)
cmetal= (2173.8717)/(1.43241)
cmetal= 1517.63
To find the percentage error for our specific heat we use the following formula:
|Accepted Value – Our Value | / (Accepted Value * 100)
|380 – 1517.63| / 380 * 100
1137.63/38000
299.38%
The error for Aluminum on our first trial is 299.38%
To find the uncertainties in our measurement we reorder the following formula Error Q / q = (Error a/a + Error b/b…) to Error Q = (Error a/a + Error b/b…)*q. Because we used cmetal= (mh2o*ch2o*(Ti-Te)h2o)/(mmetal*(Ti-Te)metal) we need to add the estimated error of Te and Ti and use the formula above for the multiplications.
cmetal= (mh2o*(Ti-Te)h2o)/(mmetal*(Ti-Te)metal) – We ignore the specific heat of H2O because no error is present.
cmetal= (0 .005* 0.10)/( 0.005*0.10) – Our Error Values
1517.63= (.40005*1.30)/(.01795*79.80) - Our Results
Uncertainty = (0.005/.40005 + 0.10/1.30 + 0.005/.01795 + 0.10/79.80)1517.63
Uncertainty = 560.3487
The result is 1517.63 ± 600. Hence our range is from 917.63 to 2117.63
Trial 2 – Aluminum
To find the specific heat of Aluminum we have to reorder mh2och2o(Ti-Te)metal=mmetalcmetal(Ti-Te)metal to:
cmetal= (mh2o*ch2o*(Ti-Te)h2o)/(mmetal*(Ti-Te)metal)
cmetal= (.404*4180*1.30)/(.01795*79.90)
cmetal= (2195.336)/(1.434205)
cmetal= 1550.10
To find the percentage error for our specific heat we use the following formula:
|Accepted Value – Our Value | / (Accepted Value * 100)
|380 – 1550.10| / 380 * 100
1170.1/38000
307.92%
The error for Aluminum on our first trial is 307.92%
To find the uncertainties in our measurement we reorder the following formula Error Q / q = (Error a/a + Error b/b…) to Error Q = (Error a/a + Error b/b…)*q. Because we used cmetal= (mh2o*ch2o*(Ti-Te)h2o)/(mmetal*(Ti-Te)metal) we need to add the estimated error of Te and Ti and use the formula above for the multiplications.
cmetal= (mh2o*(Ti-Te)h2o)/(mmetal*(Ti-Te)metal) – We ignore the specific heat of H2O because no error is present.
cmetal= (0 .005* 0.10)/( 0.005*0.10) – Our Error Values
1550.10= (.404*1.30)/(.01795*79.90) - Our Results
Uncertainty = (0.005/.404 + 0.10/1.30 + 0.005/.01795 + 0.10/79.90)1550.10
Uncertainty = 572.1456
The result is 1550.10 ± 600. Hence our range is from 950.10 to 2150.10
CONCLUSIONS & EVALUATION:
After analyzing the data I can conclude that Aluminum had the highest specific heat, because of its mass. The specific heat of aluminum is almost double of that of brass or steel even thou the temperature change in the water where similar for all the metals. This is because the mass of our metal has a greater impact then the other values in our formula that calculates the specific heat.
Over all we had an error of 174.06% that was due to the following reasons:
- The initial temperature of the metal caused an error because of the instrument we used to measure the temperature. Instead of measuring the temperature of the metal we measured the temperature of the water; hence the temperature of the metal differed from what we used as an initial temperature.
- Another error could have occurred when measuring the initial and final temperature of the water.
- The main source of error is the apparatus used, which unveiled itself to be precise but not accurate. With this I mean that the recorded results where very similar to each other, but not correct. An example, in both the brass trials that where conducted, I calculated an uncertainty of ± 100, Hence our results where precise but not correct.
- Another source of error was that on one of our trials we did not use the same materials, but we changed them, hence this can explain why our second trial had a greater error then our first.
To conclude this lab helped me understand how transfer of heat works, how certain materials transfer more heat then other; such as aluminum, which is the main reason why we find it everywhere in our kitchen. In this lab we had a great error percentage which was mostly caused by our apparatus, even thou our results was precise.
Form created by Gary Allan
108403.doc
