Respuesta :
Solution :
A). As the time passes the temperature of the coffee tends to acquire the temperature of the room, so the limiting value of the temperature of the coffee is 20°C. i.e.
[tex]$\lim_{t \to \infty} T(t)=20$[/tex]
B). And limiting value of rate of cooling is given by :
[tex]$\lim_{t \to \infty} \ \frac{dT}{dt} =\lim_{t \to \infty} \ [k(T-T_{room})] $[/tex]
[tex]$=k . \lim_{t \to \infty} (T-T_{room})$[/tex]
[tex]$=k .[ \lim_{t \to \infty} T-\lim_{t \to \infty} T_{room}]$[/tex]
[tex]$=k.[T_{room}-T_{room}]$[/tex]
= k. 0
= 0
C). Given, [tex]$\frac{dT}{dt} = -1, $[/tex] when T(t) = 70° using this in the given equation,
-1 = k.(70-20)
k = -0.02
D). By Euler method, we get
[tex]$T_{n+1}=T_n + h \ f(t_n, T_n)$[/tex]
[tex]$t_{n+1}=t_n +h$[/tex]
where, [tex]$f(t,T) =k(T-T_{room})$[/tex]
= -0.02(T - 20)
We have [tex]$T_0 = 90^\circ$[/tex] at t = 0 and h = 2.
So [tex]$t_1 = 0+2 = 2$[/tex]
∴ [tex]$T_1=T_0 + h \ f(t_0,T_0)$[/tex]
= 90+2[-0.02(90-20)]
= 87.2
At [tex]$t_2 = 2+2 = 4$[/tex]
[tex]$T_2=T_1 + h \ f(t_1,T_1)$[/tex]
= 87.2+2[-0.02(87.2-20)]
= 84.51
At [tex]$t_3 = t_2+2 = 4+2=6$[/tex]
[tex]$T_3=T_2 + h \ f(t_2,T_2)$[/tex]
= 84.51+2[-0.02(84.51-20)]
= 81.93
At [tex]$t_4 = t_3+4 = 6+2=8$[/tex]
[tex]$T_4=T_3 + h \ f(t_4,T_4)$[/tex]
= 81.93+2[-0.02(81.93-20)]
= 79.45
At [tex]$t_5 = t_4+2 = 8+2=10$[/tex]
[tex]$T_5=T_4 + h \ f(t_4,T_4)$[/tex]
= 79.45+2[-0.02(79.45-20)]
= 77.07
So after 10 minutes, the temperature of the coffee will be 77.07°C.
