Answer:
The possible equation is; T(t) = 20 + 80·![e^{(-0.094 \cdot t)}](https://tex.z-dn.net/?f=e%5E%7B%28-0.094%20%5Ccdot%20t%29%7D)
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;
![T(t) = T_s + (T_0 - T_s) \cdot e^{k \cdot t}](https://tex.z-dn.net/?f=T%28t%29%20%3D%20T_s%20%2B%20%28T_0%20-%20T_s%29%20%5Ccdot%20e%5E%7Bk%20%5Ccdot%20t%7D)
![\dfrac{dT}{dt} = -k(T - T_a)](https://tex.z-dn.net/?f=%5Cdfrac%7BdT%7D%7Bdt%7D%20%3D%20-k%28T%20-%20T_a%29)
Where;
T(t) = The temperature after time, t =
=The room temperature = 20°C
T₀ = The initial temperature of the tea = 100°C
k = Constant
T(5) = 70°C, therefore, we have;
![70 = 20 + (100 - 20) \cdot e^{k \cdot 5}](https://tex.z-dn.net/?f=70%20%3D%2020%20%2B%20%28100%20-%2020%29%20%5Ccdot%20e%5E%7Bk%20%5Ccdot%205%7D)
50/80 = ![e^{k \cdot 5}](https://tex.z-dn.net/?f=e%5E%7Bk%20%5Ccdot%205%7D)
5·k = ㏑(5/8)
k = ㏑(5/8)/5 ≈ -0.094
Therefore, the possible equation is given as follows;
T(t) = 20 + 80·![e^{(-0.094 \cdot t)}](https://tex.z-dn.net/?f=e%5E%7B%28-0.094%20%5Ccdot%20t%29%7D)
The asymptotes are Lim(t → ∞), T → 20°C and Lim(t → 0) T → 100°C