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

[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.

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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​

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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