Question

You place a cup of 200oF coffee on a table in a room that is 67oF, and 10 minutes later, it is 195oF. Approximately how long will it be before the coffee is 180oF? Use Newton's law of cooling:

Answer 1

The answer is 43 minutes APEXX

Answer 2

Answer:

After 50.4 min.

Step-by-step explanation:

Newton's law of cooling-

$T(t)=t_a+(t_0-t_a)e^{-kt}$

where,

$t_0$ = the initial temp. = 200° F

k = 0.1947,

$T(t)$ = 180° F

$t_a$ = 67° F

Putting the values,

$\Rightarrow 180=67+(200-67)e^{-0.1947\cdot t}$

$\Rightarrow 180-67=(200-67)e^{-0.1947\cdot t}$

$\Rightarrow (200-67)e^{-0.1947\cdot t}=180-67$

$\Rightarrow 133e^{-0.1947\cdot t}=113$

$\Rightarrow e^{-0.1947\cdot t}=\dfrac{113}{133}$

$\Rightarrow \ln e^{-0.1947\cdot t}=\ln \dfrac{113}{133}$

$\Rightarrow -0.1947\cdot t=\ln \dfrac{113}{133}$

$\Rightarrow t=\dfrac{\ln \dfrac{113}{133}}{-0.1947}=0.84\ h=50.4\ min$