Question

A cup of coffee with temperature 155degreesF is placed in a freezer with temperature 0degreesF. After 15 ​minutes, the temperature of the coffee is 46degreesF. Use​ Newton's Law of Cooling to find the​ coffee's temperature after 20 minutes.

Answer 1

= 30.67°F

≅ 30.70°F

Step-by-step explanation:

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

`46 = 0+(155-0)e^(-k* 15)46 = 155e^(-15k)`

Now take natural log on both the sides

`ln(46)=ln(155e^(-15k))ln(46)=ln(155)+ln(e^(-15k))ln(46)=ln(155)=-15k\\3.8286 - 5.0434 = -15k`

k = 0.081

`T(t)=0+(155-0)e^((0.081* 20))`

`= 155e^(-1.62)`

= 155 (0.1979)

= 30.67°F

≅ 30.70°F