3 years ago

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Mathematics

1 answer:

JulsSmile [24]3 years ago

5 0

Answer:

.- .-. . / -.-- --- ..- / .-. . .- .-.. .-.. -.-- / - .... .- - / -... --- .-. . -.. ..--..

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|a|>b
assume
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|t+4|>10
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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 temperat

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Answer:

21.68 minutes ≈ 21.7 minutes

Step-by-step explanation:

Given:

$T=60+40e^{kt}$

Initial temperature

T = 100°C

Final temperature = 60°C

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

Now,

using the given equation

$T=60+40e^{kt}$

at T = 90°C and t = 3 minutes

$90=60+40e^{k(3)}$

$30=40e^{3k}$

$e^{3k}=\frac{3}{4}$

taking the natural log both sides, we get

3k = $\ln(\frac{3}{4})$

3k = -0.2876

k = -0.09589

Therefore,

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

$65=60+40e^{( -0.09589)t}$

$5=40e^{( -0.09589)t}$

$e^{( -0.09589)t}=\frac{5}{40}$

$e^{( -0.09589)t}=0.125$

taking the natural log both the sides, we get

( -0.09589)t = ln(0.125)

( -0.09589)t = -2.0794

t = 21.68 minutes ≈ 21.7 minutes

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