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Yanka [14]
3 years ago
10

Help plz this is part b to my last question​

Chemistry
1 answer:
Elden [556K]3 years ago
3 0

Answer:

% error Aluminum: 0.34%

% error Steel: 16.67%

% error Lead: 23.08%

Explanation: Given our observed values of specific heat, to find the % error, we must know the true values as well. % error can be defined as, % error = |observed value - true value | / true value.

Our given data to solve this problem is as follows:

Aluminum Specific Heat Observed Value: 0.9 (given), Aluminum Specific Heat True Value: 0.897 (research).

Steel Specific Heat Observed Value: 0.49 (given), Steel Specific Heat True Value: 0.42 (research).

Lead Specific Heat Observed Value: 0.16 (given), Lead Specific Heat True Value: 0.13 (research).

___________

Since we have all the information, just subtract the true value from the observed value, and divide by the true value.

Aluminum Specific Heat % error:

observed value - true value / true value →

0.9 - 0.897 / 0.897 = 0.003 / 0.897 = 1/299 ≈ 0.34 %

Steel Specific Heat % error:

observed value - true value / true value →

0.49 - 0.42 / 0.42 = 0.07 / 0.42 = 1/6 ≈ 16.67%

Lead Specific Heat % error:

observed value - true value / true value →

0.16 - 0.13 / 0.13 = 0.03 / 0.13 = 3/13 ≈ 23.08%

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Answer:

A. releases a large amount of heat

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