Question

question 8 of 10
you place a cup of 210°f coffee on a table in a room that is 68°f, and 10 minutes later, it is 200°f. approximately how long will it be before the coffee is 180°f? use newtons law of cooling:
t(t)=t_a+(t_o - t_a)e^{-kt}
a. 15 minutes
b. 33 minutes
c. 45 minutes
d. 1 hour

Explanation:

Step1: Identify the given values

$T_0 = 210^{\circ}F$, $T_A=68^{\circ}F$, at $t = 10$ minutes, $T(10)=200^{\circ}F$. First, find the value of $k$.
Substitute into $T(t)=T_A+(T_0 - T_A)e^{-kt}$:
$200 = 68+(210 - 68)e^{-10k}$
$200-68=(210 - 68)e^{-10k}$
$132 = 142e^{-10k}$
$e^{-10k}=\frac{132}{142}$
Take the natural - logarithm of both sides:
$-10k=\ln(\frac{132}{142})$
$k=-\frac{1}{10}\ln(\frac{132}{142})\approx-\frac{1}{10}\ln(0.9296)\approx\frac{1}{10}\times0.073\approx0.0073$

Step2: Find the time $t$ when $T(t) = 180^{\circ}F$

Substitute $T(t)=180$, $T_0 = 210$, $T_A = 68$, and $k\approx0.0073$ into $T(t)=T_A+(T_0 - T_A)e^{-kt}$:
$180=68+(210 - 68)e^{-0.0073t}$
$180 - 68=(210 - 68)e^{-0.0073t}$
$112 = 142e^{-0.0073t}$
$e^{-0.0073t}=\frac{112}{142}\approx0.7887$
Take the natural - logarithm of both sides:
$-0.0073t=\ln(0.7887)$
$t=-\frac{\ln(0.7887)}{0.0073}\approx-\frac{- 0.237}{0.0073}\approx32.47\approx33$ minutes

Answer:

B. 33 minutes