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(a) A cup of coffee has temperature 95 $ ^\circ C $ and takes 30 minutes to cool to 61 $ ^\circ C $ in a room with temperature 20 $ ^\circ C $. Use Newton's Law of Cooling (section 3.8) to show that the temperature of the coffee after $ t $ minutes is

$$ T(t) = 20 + 75 e^{-kt} $$

where $ k \approx 0.02 $.

(b) What is the average temperature of the coffee during the first half hour?

(A). $T(t)=20+75 e^{-0.02 t}$

(B). $76.4^{\circ} C$

Applications of Integration

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Okay. We know they've given us an equation. And we know now that if capital T is the temperature of a coffee after time teeth, then we know using Newton's law of cooling, which is listed in this chapter. We know we have duty over TFT minus 20 and we know the integral of this is equivalent to and to roll of negative K. Did you? Which gives us cheat to the T minus 20. Even C E to the negative. Katie could remember natural log times e we end up lifting up the exponents. Therefore, we have capital. Chief Chief is 20 co c e to the negative. Kate. Cheap number K is your point. You're too and t is zero there for capital to use 95 see if 75 therefore capital t of tea, the equation would be 20 with 75 e to the negative 0.2 times. T, this is part A. Now we're moving on to part B of this question. We know we're gonna be using the average value Formula one over B minus. Eso won over 30 minutes. You're from A to B from 0 30 capital G of tea. Do you see? We know what our function is. We just figured this out in part, right? Snowy, canned. Hello again. I'm plugging in the function divide by the constant care. When we take the integral from 0 to 30 this is equivalent to 76.4 degrees Celsius upon plugging in.