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

Answer:

45.50° F

Step-by-step 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

$103=0+(155-0)e^{-k\times 5}$

$103=155e^{-5k}$

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

$k=\frac{0.40872}{5}$

k = 0.0817

$T(t)=0+(155-0)e^{(0.0817\times 15)}$

   = $155e^{-1.226}$

   = 155 (02935)

  = 45.49 ≈ 45.50° F