Question

dt Newton's law of cooling states that the rate of change in the temperature (t) of a body is proportional to the difference between the temperature of the medium M(t) and the dT temperature of the body. That is, = K[M(1) – TCC), where is a constant. Let K = 0.03 (min) and the temperature of the medium be constant, m(t) = 295 kelvins. If the body is initially at 364 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. (a) The temperature of the body after 30 minutes is 323.01 kelvins. (Round to two decimal places as needed.) (b) The temperature of the body after 60 minutes is kelvins. (Round to two decimal places as needed.)

Answer 1

given that -* [met) - T)] K=0.03 Mit) = 295K , T100=364K dT at here so dr 0.03[295-T] dt Euler's method states that Ynto = In thi f(th, Yon) we have that h= 5, too, Yo = 364 so ti = toth 55 414) = y15) = 364+h. f(0,364) = 364+5 18.85) = 408.25 9'te=5+5310 Y(10) = 408.25thit (5,608.25 ) = 408.25 +5.18.7) = 451-75 +3 = lots=15 = 494.5 & yltz) = y 10) = 451.75 +5.18.55). t4 = 15+5=20 494.5 → Y (+4) = 4120) = 536-5+ 5(8.4) 57775 to= rots » Y(ts) = 4125] 536.5 + 5(8.25) = 577.75 to = 25+5= 30 - 25 /ylto) 9130) = 597.75+ 51 8:1) = 618.25 4 (30) 618, 25k 47= 30ts =35 Y() = (35) = 618.25+ 517.45) = 658 t8= 35+5= 40 » YIts) = 4(60) = 658+5(7.8) = 697 to = 4045 = 45 4(45) = = y ltq) = 697 +51 7.65) = 735.25