Question

The rate of cooling of a body can be expressed as dT dt :-k(T-T) where T = temperature of the body (°C), Ta= temperature of the surrounding medium (°C), and k=a proportionality constant (per minute). Thus, this equation (called Newton's law of cooling) specifies that the rate of cooling is proportional to the difference in the temperatures of the body and of the surrounding medium. If a metal ball heated to 80 °C is dropped into a lake where the temperature is Ta, the temperature of the ball changes as in Table 3-1. Table 3-1 Time, min T, °C 0 80 10 44.5 20 30.0 30 24.1 40 21.7 50 20.7 The visual representation of the cooling effect is presented in Figure 3-1. T Figure 3-1 Apply a suitable regression model to represent the rate of cooling of the metal ball. Show how mathematical constant(s) are determined in the regression. Make logical assumption(s) for the temperature of the lake. [15 marks]

Answer 1

given rate of cooling of body in ake. dT -k(T-Ta) dt T: Temperature of body (°C) medium (C) surroun. unding Ta= tumperature of constant ( min ) k= proportionality at t-0, To: 80'c given suitable regression model to represent the Apply the ball rate woling of of Tas 20°C Am) Ansume method to find using forward difference

dT . (**), 1 -1) +41 (412) -3 110) dt ah ha stup nize = 10 mini t-o at 20;-0 -|(__) +4 (x,) - 34(40) . d'o) 2h 3* 80 J'(o)- 30 + 4×445 2x10 -4.6 °C/min - ponse difference centered use ( 1 ) ( x ) - }(2-) ah (2)

I't-10) 30 - 80 -2.5 °C/min 2xio 24.1- 44.5 -1.02 "C/min I'lt: 20) 2 YIO 21.7-30 -0.415 "C/min I'(4:30) - 2 XD -0.17 °C/min. 2 20.7.-24.1 I'CE:40) 2x10 method: difference backward use f'(25) = 3 4(X;)- 48 («,-) + +(21-2) ah d' t; 50) 31(t. 50)-41 (t=40)+1(5:20) ah 13

I'lts 50) 2 3* 20.7 - 4* 21.7 + 24.1 = -0.03'c min axio have the following values: - Now Ta J-Ta dT dt 1 t 60 -4.6 0 80 20 2.5 20 24.5 44.5 30 20 20 -1.02 30 24.1 ao -0.415 40 21.7 20 1.7 -0.17 20 0.7 50 20,7 -0.03

Now can plot excel. on ok( T - Ta) dT dt f on y-axis with dT dt x-axis. On CT-Ta) line (y=mx) a straight this represente linear regression do O can origin, we through the value of the trendline is Our slope of 'K'. trendline ( linear regression), On adding get the cote with -0.077 x 0.155 +12). R? 0.981 co-efficient of determination

which indicates good fit. and from trendline, the equation of -0.077 nlope ka 0.077 I min (Exel plot below)

Excel plot:

0 10 20 30 40 50 60 70 -0.5 -1 -1.5 f(x) = - 0.077x -0.155 R2 = 0.981 -2 dT/dt -25 -3 -3.5 -4 -4.5 -5 T-Ta

----*-----