Question

4. A ball with initial temperature of 1200K is allowed to cool down in air at an ambient temperature of 300K. Assuming heat is lost only due to radiation, the differential equation for the temperature of the ball is given by db = -2.2067 x 10" (69-81 x 10').6(0) = 1200 K By using Euler's method with a step size of h = 60 seconds, show that the temperature at t = 480 seconds is about 615K.

engineering math question.. please answer all by showing step by step calculation clearly and easy to read and understand.. thanks

Answer 1

Doubt in any step then comment below. I will explain you.

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here i use T for temperature...because to write that typical sign is so irritating okk...so you not confuse ... T stands for temperature ..  

-12 A dI 22.2067 x 10 (TY-81x108) da flr h e Go Tot 1200 Method Euler f(T) Titia Tith Titia Tit cool (-2. 20674 10t? (9 - 81x108) Ti - 1.32402 x 101 (T" - 814108) / Titi = 200) - 81xlu To= 1200 1.32402x10 to $ 1200 - E 1200 - 132402 Xp070 (2,0655 XI.!?) - 1200 - 293, 47633) 926.523669

Now we want t=480 ...

So we have to calculate T_8 ...

81 vi8) т. е т . 1, 31 ч. 2 ч. (Т. - - 96. 366 5 - 96. Чуха) 83 о, от 1 до - g 1ч.3) Тет. - . go гч --- ( т.ч. 2 ч. 8) Т. 3 ч. 2 что - 13. Т. 2 P3o. or - с. 776 9 2 81 6 (Ч 2278 169. - Чs.oъ2 o 731 2 78 1 ч . 26 8. 2 12 7, 20 Ч. 2.7 3S 1ч666 і 16 2, ч. 23 - 12 3 26 Ч. 23 - 688. ост?? ? 68. окт 23 - 29. 62 76 722 = 6s 9, Чт 4 6 87 Т? 6 7Чуч6 г. 23.441 6 3s 98 705 Т. 63s чx 715 - 20 ( 9 67 січ. 966 ~1615k