Question

Suppose you have just poured a cup of freshly brewed coffee with temperature 90∘C in a room where the temperature is 20∘C.

Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. Therefore, the temperature of the coffee, T(t), satisfies the differential equation

dT/dt=k(T−Troom)

where Troom=20 is the room temperature, and k is some constant.
Suppose it is known that the coffee cools at a rate of 1∘C per minute when its temperature is 70∘C.

A. What is the limiting value of the temperature of the coffee?
limt→[infinity]T(t)= ________
B. What is the limiting value of the rate of cooling?
limt→[infinity]dT/dt= _________
C. Find the constant k in the differential equation.
k= _________
D. Use Euler's method with step size h=2 minutes to estimate the temperature of the coffee after 10 minutes.
T(10)= _________

Answer 1

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.

`$\lim_(t \to \infty) T(t)=20$`

B). And limiting value of rate of cooling is given by :

`$\lim_(t \to \infty) \ (dT)/(dt) =\lim_(t \to \infty) \ [k(T-T_(room))] $`

`$=k . \lim_(t \to \infty) (T-T_(room))$`

`$=k .[ \lim_(t \to \infty) T-\lim_(t \to \infty) T_(room)]$`

`$=k.[T_(room)-T_(room)]$`

= k. 0

= 0

C). Given,
`$(dT)/(dt) = -1, $` when T(t) = 70° using this in the given equation,

-1 = k.(70-20)

k = -0.02

D). By Euler method, we get

`$T_(n+1)=T_n + h \ f(t_n, T_n)$`

`$t_(n+1)=t_n +h$`

where,
`$f(t,T) =k(T-T_(room))$`

= -0.02(T - 20)

We have
`$T_0 = 90^\circ$` at t = 0 and h = 2.

So
`$t_1 = 0+2 = 2$`

∴
`$T_1=T_0 + h \ f(t_0,T_0)$`

= 90+2[-0.02(90-20)]

= 87.2

At
`$t_2 = 2+2 = 4$`

`$T_2=T_1 + h \ f(t_1,T_1)$`

= 87.2+2[-0.02(87.2-20)]

= 84.51

At
`$t_3 = t_2+2 = 4+2=6$`

`$T_3=T_2 + h \ f(t_2,T_2)$`

= 84.51+2[-0.02(84.51-20)]

= 81.93

At
`$t_4 = t_3+4 = 6+2=8$`

`$T_4=T_3 + h \ f(t_4,T_4)$`

= 81.93+2[-0.02(81.93-20)]

= 79.45

At
`$t_5 = t_4+2 = 8+2=10$`

`$T_5=T_4 + h \ f(t_4,T_4)$`

= 79.45+2[-0.02(79.45-20)]

= 77.07

So after 10 minutes, the temperature of the coffee will be 77.07°C.