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Unformatted text preview: mperature difference is so relatively large, between the cooked turkey and the room air, the cooling rate is also relatively great. As time passes, the temperature difference decreases and therefore the cooling rate slows. Eventually, if left unattended, the cooling rate decreases to 0 because the turkey’s temperature will have cooled to room temperature. The turkey can’t get any cooler than the air around it. We can generate a differential equation to model this phenomenon. dT
= K (T ! A) dt where dT/dt represents the rate of change of the object’s temperature (T) with respect to time (t), with constant of proportionality K and ambient temperature A. We can use separation of variables to solve this differential equation, formulating a general solution that expresses the obje...
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 Summer '09
 Calculus, Equations

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