Question

A pot of stew is placed on a stove to heat. The temperature of the liquid reaches 170°F, and then the pot is taken off the burner and placed on a kitchen counter. The temperature of the air in the kitchen is 76°F.
If k = 0.34, the temperature of the liquid after 7 hours will be

ºF. (Hint: Use Newton’s cooling model.)

Answer 1

Answer:

$T =84.7\°F$

Step-by-step explanation:

Newton's cooling model is:

$T(t) =T_0 + (T_i-T_0)e^{-kt}$

Where

t is the time in hours

$T_0$ is the ambient temperature

$T_i$ is the initial temperature of the body

$k=0.34$

We know that:

$T_0=76\°F$

$T_i=170\°F$

Then after 7 hours the temperature of the liquid will be:

$T(7) =76 + (170-76)e^{-0.34(7)}$

$T =84.7\°F$