Question

A cup of coffee with temperature 155degreesF is placed in a freezer with temperature 0degreesF. After 5 ​minutes, the temperature of the coffee is 103degreesF. Use​ Newton's Law of Cooling to find the​ coffee's temperature after 15 minutes.

Answer 1

45.50° F

Explanation:

As per Newton's law,

`T(t) = T_(s)+(T_(0) + T_(s))e^(-kt)`

When T(t) is the final temperature

`T_(s)` = Temperature of surrounding

`T_(0)` = Initial temperature

t =duration of cooling

k = constant

`103=0+(155-0)e^(-k* 5)`

`103=155e^(-5k)`

Now take natural log on both the sides

`ln(103)=ln(155e^(-5k))`

`ln(103)=ln(155)+ln(e^(-5k))`

`ln(103)=ln(155)=-5k`

4.6347 - 5.0434 = -5k

`k=(0.40872)/(5)`

k = 0.0817

`T(t)=0+(155-0)e^((0.0817* 15))`

=
`155e^(-1.226)`

= 155 (02935)

= 45.49 ≈ 45.50° F