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

After 50.4 min.

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)=(113)/(133)`

`\Rightarrow \ln e^(-0.1947\cdot t)=\ln (113)/(133)`

`\Rightarrow -0.1947\cdot t=\ln (113)/(133)`

`\Rightarrow t=(\ln (113)/(133))/(-0.1947)=0.84\ h=50.4\ min`