Question

question 6 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. 45 minutes
b. 33 minutes
c. 1 hour
d. 15 minutes

Explanation:

Step1: Identify known values

$T_A=68$, $T_0=210$, $t=10$ when $T=200$

Step2: Solve for k

Substitute into formula: $200=68+(210-68)e^{-10k}$
Simplify: $132=142e^{-10k}$ → $e^{-10k}=\frac{132}{142}\approx0.9296$
Take ln: $-10k=\ln(0.9296)\approx-0.0729$ → $k\approx0.00729$

Step3: Find t when T=180

Substitute: $180=68+142e^{-0.00729t}$
Simplify: $112=142e^{-0.00729t}$ → $e^{-0.00729t}=\frac{112}{142}\approx0.7887$
Take ln: $-0.00729t=\ln(0.7887)\approx-0.237$ → $t\approx\frac{0.237}{0.00729}\approx32.5\approx33$

Answer:

B. 33 minutes