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Contact [7]
3 years ago
15

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​

Mathematics
1 answer:
blsea [12.9K]3 years ago
4 0

9514 1404 393

Answer:

  5.8496 min

Step-by-step explanation:

  \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}}

__

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

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