Question

Heat Transfer 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.

Short answer

The time required for the object to cool from 300°F to 250°F is given by the calculation \(t = \frac{10}{\ln 2} \cdot \ln \left(\frac{200}{150}\right)\) minutes.

Step-by-step solution

Understand the Formula

The cooling of an object is described by the equation \(t = \frac{10}{\ln 2} \int_{T1}^{T2} \frac{1}{T - 100} dT\), where \(t\) is the cooling time, \(T1\) and \(T2\) are the initial and final temperatures in Fahrenheit, and the integral represents the change in temperature over time.

Substitute the values

Set up the integral to find the time required for the object to cool from 300°F to 250°F. The integral is thus \(\frac{10}{\ln 2} \int_{250}^{300} \frac{1}{T - 100} dT\).

Evaluate the Integral

The antiderivative of \(\frac{1}{T - 100}\) is \(\ln |T - 100|\). Applying the fundamental theorem of calculus, we evaluate the antiderivative at the upper and lower limits of integration: \(\ln |300 - 100| - \ln |250 - 100|\). Simplifying, we get \(\ln \left(\frac{200}{150}\right)\).

Final Calculation

Multiply this value by the constant at the front of the integral. This gives the time \(t\) in minutes: \(t = \frac{10}{\ln 2} \cdot \ln \left(\frac{200}{150}\right)\).