Question

after heating up in a teapot, a cup of hot water is poured at a temperature of 208°f. the cup sits to cool in a room at a temperature of 67°f. newtons law of cooling explains that the temperature of the cup of water will decrease proportionally to the difference between the temperature of the water and the temperature of the room, as given by the formula below: (t = t_{a}+(t_{0}-t_{a})e^{-kt}) (t_{a}=) the temperature surrounding the object (t_{0}=) the initial temperature of the object (t =) the time in minutes (t=) the temperature of the object after (t) minutes (k =) decay constant the cup of water reaches the temperature of 191°f after 3 minutes. using this information, find the value of (k), to the nearest thousandth. use the resulting equation to determine the fahrenheit temperature of the cup of water, to the nearest degree, after 5.5 minutes. enter only the final temperature into the input box.

Explanation:

Step1: Identify given values

$T_a = 67^{\circ}F$, $T_0=208^{\circ}F$, $T = 191^{\circ}F$, $t = 3$ minutes. Substitute into $T = T_a+(T_0 - T_a)e^{-kt}$.
$191=67+(208 - 67)e^{-3k}$

Step2: Simplify the equation

First, simplify the right - hand side:
$191=67 + 141e^{-3k}$
Subtract 67 from both sides:
$191-67=141e^{-3k}$
$124 = 141e^{-3k}$
Then, divide both sides by 141:
$\frac{124}{141}=e^{-3k}$

Step3: Take the natural logarithm of both sides

$\ln(\frac{124}{141})=\ln(e^{-3k})$
Since $\ln(e^{-3k})=-3k$, we have:
$\ln(\frac{124}{141})=-3k$
$k=-\frac{\ln(\frac{124}{141})}{3}$
$k\approx-\frac{\ln(0.88)}{3}\approx-\frac{- 0.1278}{3}\approx0.043$

Step4: Find the temperature at $t = 5.5$ minutes

Now that $k\approx0.043$, $T_a = 67^{\circ}F$, $T_0 = 208^{\circ}F$, and $t = 5.5$ minutes. Substitute into $T = T_a+(T_0 - T_a)e^{-kt}$:
$T=67+(208 - 67)e^{-0.043\times5.5}$
$T=67 + 141e^{-0.2365}$
Since $e^{-0.2365}\approx0.789$, then:
$T=67+141\times0.789$
$T=67 + 111.249\approx178^{\circ}F$

Answer:

$178$