Using Newton's law of cooling, time taken by the coffee to cool down to 180°F from the initial temperature of 210°F is 33 minutes approximately.
Newton's law of cooling states that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its environment.
Let θ₀ be the temperature of the surroundings = 68°F
Consider a cooling from 210°F to 200°F
Let θ₁ be the initial temperature = 210°F
Let θ₂ be the final temperature = 200°F
Time taken = t = 10 minutes
By Newton's law of cooling,
[tex]\frac{d\theta}{dt} = C(\theta - \theta_{0})\\\\\frac{\theta 1 - \theta 2}{dt} = C(\frac{\theta 1 + \theta 2}{2} - \theta_{0})\\\\\frac{210 - 200}{10} = C(\frac{210+200}{2} - 68)\\ \\ 1 = C(137)\\\\C = \frac{1}{137}[/tex]
where C is the rate of cooling.
Consider a cooling from 200°F to 180°F
Let θ₁ be the initial temperature = 200°F
Let θ₂ be the final temperature = 180°F
Time taken = t = ? minutes
[tex]\frac{d\theta}{dt} = C(\theta - \theta_{0})\\\\\frac{\theta 1 - \theta 2}{dt} = C(\frac{\theta 1 + \theta 2}{2} - \theta_{0})\\\\\frac{200 - 180}{t} = \frac{1}{137} (\frac{200+180}{2} - 68)\\ \\ \frac{20}{t} = 0.890510949\\\\t = \frac{20}{0.890510949} = 22.45901639[/tex]
The time taken by the coffee to cool down from 200°F to 180°F is 23 minutes approximately.
Total time taken by the coffee to cool down from 210°F to 180°F is (23 + 10) = 33 minutes approximately.
Learn more about Newton's law of cooling here
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