(Newton's Law of Cooling.) Let T(t) be the temperature of an object in a room of fixed ambienttemperature A. The rate of change of temperature is proportional to the difference between T and A,i.e., T satisfiesdTdt-k(T - A).(a) Find a general solution for T(t).(b) An iced tea at 5°C is placed in a cabana where the temperature is 35°C. After ten minutes theiced tea is at 10°C. What is the temperature of the drink after 40 minutes?(c) What happens as too in this model?

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Answered: Newton's Law of Cooling.) Let T(t) be…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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The displacement s (in metres) of a particle as a function of time t (in seconds) is given by s(t) = V 4t2 – 11· Find the velocity of this particle at t = 3 seconds.
c) A cup of water with a temperature of 90°C temperature of 24° C. After 15 minutes, the water temperature is 55° C. Newton's Law of placed in a room with a constant dT cooling states that the rate at which an object cools is =Dk(T-To) , where k is the dt %3| constant of proportionaility, T is the temperature of the object, and To is the surrounding temperature. Use the Newton's Law to calculate the time for the water temperature to reach 30° C.
A piece of chicken is boiled until its internal temperature reaches 165 °F. The chicken is then transferred to a refrigerator held at 49 °F to cool down. The internal temperature of the chicken at time t is given by T(t), and the temperature obeys Newton's law dT = -k(T - Tm) where Tm is the temperature of the refrigerator and k > 0 is a constant. After 5 minutes in the refrigerator, the internal temperature of the chicken is 155 °F. How long will it take for the temperature of the chicken to reach 70 °F?
2. (Separable and Newton's Law of Cooling) Let T(t) be the temperature of the room and T, the outside temperature (assumed to be constant in Newton's law). Then by Newton's law dt k(T – T,). In this problem, the rate at which a substance cools in air is directly proportional to the difference between the temperatures of the substance and that of air. The temperature of air, TA, is 30° and the substance cools from 100° to 70° in 15 minutes. How long does it take for the substance to cool from 100 to 50"? A. 45.30 minutes B. 43.60 minutes C. 35.39 minutes D. 33.59 minutes Solution. Hints: 1. Find the general solution of the equation = k(T – T,) by separable. 2. Let t = 0 and T = 100° to find C. 3. Let t = 15 and T = 70 to find k. 4. Let T = 50° to find t.
Newton's law of cooling states that the rate of change in the temperature T(t) of a body is proportional to the difference between the temperature of the medium M(t) dT 1 and the temperature of the body. That is, = K[M(t) - T(t)], where K is a constant. Let K = 0.04 (min) and the temperature of the medium be constant, dt M(t) = 292 kelvins. If the body is initially at 360 kelvins, use Euler's method with h = 0.1 min to approximate the temperature of the body after (a) 30 minutes and (b) 60 minutes.
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(1) A particle has an initial displacement of s = 30 metres when t = 0. The velocity of the particle is v= 6t² - 216t+ 1944 m/s (t > 0). (a) Find the displacement of the particle at time t. (b) Find the acceleration of the particle at time t. (c) What is the acceleration and displacement when v = 0?

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