PLEASE SHOW ALL YOUR WORK! 9.3 Graphical and Numerical Methods: Problem 6 Previo

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PLEASE SHOW ALL YOUR WORK!

9.3 Graphical and Numerical Methods: Problem 6 Previous Problem Problem List Next Problem (1 point) Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. Suppose t is time, T is the temperature of the object, and T, is the surrounding temperature. The following differential equation describes Newton's Law dT =k(T-T,), dt where k is a constant. Suppose that we consider a 95°C cup of coffee in a 18 room. Suppose it is known that the coffee cools at a rate of LC/min. when it is 70°C. Answer the following questions. 1. Find the constant k in the differential equation Answer (in per minute): k = 2. What is the limiting value of the temperature? Answer (in Celsius): T = 3. Use Euler's method with step size h = 2 minutes to estimate the temperature of the coffee after 10 minutes. Answer (in Celsius): T(10) *

Explanation / Answer

sol:

let T = temperature at any time t

the rate of change of temperature of an object is directly proportional to the difference between the object's temperature and the temperature of its surroundings.

dT/dt = k(T - 18)

at T= 70, dT/dt = -1 deg/min

-1 = k(70 - 18)

k = -1/52

dT/dt = (-1/52)(T - 18)

dT / (T - 18) = -1/52 dt

integrate both side

ln(T-18) = -1/52 t +c c is constanst

T = 18 + C e(-1/52 t)

at t=0, T =95

95 = 18 +c

77 =c

T = 18 + 77 e(-1/52 t)

c)

T(2) = T(0) + 2[(-1/52)(T(0) - 18)] = 95 + 2 (-1/52) ( 95-18) = 92.04

T(4) = T(2) + 2[(-1/52)(T(2) - 18)] = 89.19

T(6) = T(4) + 2[(-1/52)(T(4) - 18)] = 86.45
T(8) = T(6) + 2[(-1/52)(T(6) - 18)] = 83.82

T(10) = T(8) + 2[(-1/52)(T(8) - 18)]= 81.29

and so on til you get to y(10).

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