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Question

# Find the time required for an object to cool from $300^{\circ} \mathrm{F}$ to $250^{\circ} \mathrm{F}$ by evaluating$t=\frac{10}{\ln 2} \int_{250}^{300} \frac{1}{T-100} d T$where $t$ is time in minutes.

Solution

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Evaluate the given integral

\begin{aligned} t=&\dfrac{10}{\ln(2)}\bigg[\ln|T-100|\bigg]_{250}^{300}\\ =&\dfrac{10}{\ln(2)}\bigg[\ln|300-100]-\ln|250-100|\bigg]\\ =&\dfrac{10}{\ln(2)}\bigg[\ln(200)-\ln(150)\bigg]\\ =&\dfrac{10}{\ln(2)}\bigg[\ln\dfrac{200}{150}\bigg]\\ =&\dfrac{10}{\ln(2)}\ln\bigg(\dfrac{4}{3}\bigg)\approx 4.15 \text{ min} \end{aligned}

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