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Water whose temperature is at 100∘C is left to cool in a room where the temperature is 60∘C. After 3 minutes, the water temperature is 90∘. If the water temperature T is a function of time t given by T = 60 + 40 e k t T=60+40ekt, find the time for the water temperature to reach 65∘C. Round to the nearest hundredth of a minute.

Respuesta :

Answer:

21.68 minutes ≈ 21.7 minutes

Step-by-step explanation:

Given:

[tex]T=60+40e^{kt}[/tex]

Initial temperature

T = 100°C

Final temperature = 60°C

Temperature after (t = 3 minutes) = 90°C

Now,

using the given equation

[tex]T=60+40e^{kt}[/tex]

at T = 90°C and  t = 3 minutes

[tex]90=60+40e^{k(3)}[/tex]

[tex]30=40e^{3k}[/tex]

or

[tex]e^{3k}=\frac{3}{4}[/tex]

taking the natural log both sides, we get

3k = [tex]\ln(\frac{3}{4})[/tex]

or

3k = -0.2876

or

k = -0.09589

Therefore,

substituting k in 1 for time at temperature, T = 65°C

[tex]65=60+40e^{( -0.09589)t}[/tex]

or

[tex]5=40e^{( -0.09589)t}[/tex]

or

[tex]e^{( -0.09589)t}=\frac{5}{40}[/tex]

or

[tex]e^{( -0.09589)t}=0.125[/tex]

taking the natural log both the sides, we get

( -0.09589)t = ln(0.125)

or

( -0.09589)t = -2.0794

or

t = 21.68 minutes ≈ 21.7 minutes