Question

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) and the temperature of the body. That is, * = K[M(t) - T(t)], where K is a constant. Let K = 0.05 (min) and the temperature of the medium be constant, M(t) = 295 kelvins. If the body is initially at 354 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.

Answer 1

ANSWER & Gruen di (MCA)-T(1) ] and m(t)=295k . Cconstant) . Using Colex method with hor 1 min we have to find Body temperature Eles method is t aldy) (th-th. 1.) here h orto-1 dl let In our case y =) T, t- t and Cinen dt K(295-T(+)) Given To: 354k . T. Tot K(295-Tobin on T Tok (295-To) to: 1 (In = T(Axolning) so Tio To tk' (295-7o)) and T = Tit k' (295-7) subsitute Ti= Totki (295-To ) in about equation t2 = Tot k' (205-To ) + k' (ras-To-kl (2 95 To )) - Tot K( 295 To ) + k' (zar-To ) (1-K) T = Tot k' (295-To ) (2-k!)

3= It k' (295-72) substitute Tz in abue equation +z= Totki (295-To) (2-k') + ' (295-To-k" (295-To) (261)) - Tot k' (295-To) (2-*') + < (295-50) (1-k(2-x)) = Tot k' (20s-To) (2-k'+ 1 - k' (2-4')) (neglet K(2-x)) Tz - Tot k' (295-To) (3-k1) Hans from eau 0 and ( we observe Tha Tork' (295-To) (n - k') - General equation whex In Tempsatux at nxo 2 min which is still and K = kxo.1 a) Temperatus at 30 min. T300 (300x0 min -Bonin) Tan= 354+ (0.05 +0.1) (295-354) (247 005 +0:1) = 265 79k after 30 min T= 265.79k 6) Temposatux at 60 min = Too Tin= 354+ (o os * 0:2) (295-354) (549 -o os xoa) = 177. 29K after 60 min T = 177 29k.
to find the relation between Tn and T0 approximations are used by neglecting some terms as shown in the solution above