It's a math assignment. It's not a writing assignment. 4.A teacher makes a cup of coffee. At 13:00 the coffee in the cup has a temperature of 86 Celsius and isplaced in a classroom that has a constant air temperature of 21 Celsius. The teacher forgets about thecup of coffee. The coffee cools at a rate proportional to the difference between the temperature of thecoffee and the air temperature of the room. After 10 minutes, at 13:10, the temperature of the coffee is63° Celsius.(a) Let h(t) be the function that models the temperature h (in Celsius) of the coffee at time t (inminutes). Find an expression for h(t).(b) Use your model from (a) to estimate the temperature of the coffee at 13:20.

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Description: It's a math assignment. It's not a writing assignment. 4.A teacher makes a cup of coffee. At 13:00 the coffee in the cup has a temperature of 86 Celsius and isplaced in a classroom that has a constant air temperature of 21 Celsius. The teacher forget

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College Algebra

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ISBN:9781337282291

Author:Ron Larson

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Section3.5: Mathematical Modeling And Variation

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It's a math assignment. It's not a writing assignment.

4.
A teacher makes a cup of coffee. At 13:00 the coffee in the cup has a temperature of 86 Celsius and is
placed in a classroom that has a constant air temperature of 21 Celsius. The teacher forgets about the
cup of coffee. The coffee cools at a rate proportional to the difference between the temperature of the
coffee and the air temperature of the room. After 10 minutes, at 13:10, the temperature of the coffee is
63° Celsius.
(a) Let h(t) be the function that models the temperature h (in Celsius) of the coffee at time t (in
minutes). Find an expression for h(t).
(b) Use your model from (a) to estimate the temperature of the coffee at 13:20.

Expert Solution

Step 1

(a) It's given that the rate of cooling coffee is proportional to the difference between coffee temperature and room temperature. So $\frac{d h}{d t} = k (h - T_{0})$ . Where, k is a proportionality constant and $T_{0}$ is the room temperature. Here, $T_{0} = 21 ° C$ .

$\begin{array}{rcl} \frac{d h}{d t} & = & k (h - 21) \\ \frac{d h}{h - 21} & = & k d t \end{array}$

Let assume at 13:00, time $t = 0$ . So at 13:00, $t = 10 minutes$ . At $t = 0$ , $h = 86 ° C$ . Now, integrate both sides of $\frac{d h}{h - 21} = k d t$ and apply the integration limit.

$\int_{86}^{h} \frac{d h}{h - 21} = \int_{t = 0}^{t} k d t$

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