Find the time required for an object to cool from 280°F to 220°F by evaluating the following where t is time in minutes. (Round your answer to four decimal places.) *280 t= 10 In 2 1 dT T – 100 min​

Find The Time Required For An Object To Cool From 280F To 220F By Evaluating The Following Where T Is

Answers

Answer 1

9514 1404 393

Answer:

  5.8496 min

Step-by-step explanation:

  [tex]\displaystyle t=\frac{10}{\ln{2}}\int_{220}^{280}{\frac{1}{T-100}}\,dT=\frac{10(\ln{(280-100)}-\ln{(220-100)})}{\ln{2}}=\frac{10\ln{(3/2)}}{\ln{2}}\\\\\boxed{t\approx5.8496\quad\text{min}}[/tex]

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Modern calculators can perform integration with high accuracy. This one gives the same result.

Find The Time Required For An Object To Cool From 280F To 220F By Evaluating The Following Where T Is

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9514 1404 393

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9514 1404 393

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