Newton's law of cooling states that for a cooling substance with initial temperature To, the temperature T(t) after t minutes can be modeled by the equation T(t) = T, + (To – T,)e-kt, where T, is the surrounding temperature and k is the substance's cooling rate. A liquid substance is heated to 80°C. Upon being removed from the heat, it cools to 60°C in 12 min. What is the substance's cooling rate when the surrounding air temperature is 50°C? Round the answer to four decimal places. O 0.0687 O 0.0732 O 0.0813 O 0.0916

Newton's law of cooling states that for a cooling substance with initial temperature To, the temperature T(t) after t minutes can be modeled by the equation T(t) = T, + (To – T,)e-kt, where T, is the surrounding temperature and k is the substance's cooling rate. A liquid substance is heated to 80°C. Upon being removed from the heat, it cools to 60°C in 12 min. What is the substance's cooling rate when the surrounding air temperature is 50°C? Round the answer to four decimal places. O 0.0687 O 0.0732 O 0.0813 O 0.0916

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Differential Equations

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Newton's law of cooling states that for a cooling substance with initial temperature To, the temperature T(t) after t minutes can be
modeled by the equation T(t) =T, + (To – T)e-kt, where T, is the surrounding temperature and k is the substance's cooling rate.
A liquid substance is heated to 80°C. Upon being removed from the heat, it cools to 60°C in 12 min.
What is the substance's cooling rate when the surrounding air temperature is 50° C?
Round the answer to four decimal places.
O 0.0687
O 0.0732
O 0.0813
O 0.0916

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Newton’s law of cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (i.e., the temperature of its surroundings). We let (t) denote time, T(t) be the temperature of the object at time (t), (T0 = T(0)) (the object’s temperature at (t = 0)), and Tomg be the constant ambient temperature. Set u(t) = T(t) - Tomg and show from Newton’s law of cooling that (u) satisfies the initial value problem (*) u’(t) = ku(t), u(0) = T0 - Tomg for a constant (k). Solve the initial value problem (*), i.e., find u(t)) (expressed in terms of (k), T0, and Tomg . Find T(t).
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