Answer:
The possible equation is; T(t) = 20 + 80·[tex]e^{(-0.094 \cdot t)}[/tex]
The horizontal asymptotes is Lim(t → ∞), T → 20°C
The vertical asymptote is Lim(t → 0) T → 100°C
Step-by-step explanation:
The given parameters are;
The temperature of the house = 20°C
The initial temperature of the tea = 100°C
The temperature of the tea after 5 minutes = 70°C
From Newton's law of cooling, is given as follows;
[tex]T(t) = T_s + (T_0 - T_s) \cdot e^{k \cdot t}[/tex]
[tex]\dfrac{dT}{dt} = -k(T - T_a)[/tex]
Where;
T(t) = The temperature after time, t =
[tex]T_s[/tex] =The room temperature = 20°C
T₀ = The initial temperature of the tea = 100°C
k = Constant
T(5) = 70°C, therefore, we have;
[tex]70 = 20 + (100 - 20) \cdot e^{k \cdot 5}[/tex]
50/80 = [tex]e^{k \cdot 5}[/tex]
5·k = ㏑(5/8)
k = ㏑(5/8)/5 ≈ -0.094
Therefore, the possible equation is given as follows;
T(t) = 20 + 80·[tex]e^{(-0.094 \cdot t)}[/tex]
The asymptotes are Lim(t → ∞), T → 20°C and Lim(t → 0) T → 100°C