question 8 of 10\nyou place a cup of 210°f coffee on a table in a room that is 68°f, and 10 minutes later…

question 8 of 10\nyou 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:\nt(t)=t_a+(t_o - t_a)e^{-kt}\na. 15 minutes\nb. 33 minutes\nc. 45 minutes\nd. 1 hour

question 8 of 10\nyou 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:\nt(t)=t_a+(t_o - t_a)e^{-kt}\na. 15 minutes\nb. 33 minutes\nc. 45 minutes\nd. 1 hour

Answer

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