Apot 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.)

The temperature of the liquid after 7 hours will be 84.7 °F

Solution:

Initial Temperature of the liquid is 170°F:

at t=0, T(0)=170°F

Temperature of the air: Ta=76°F

k=0.34

Newton's cooling model:

T(t)=Ta+C e^(-kt)

Replacing the known values:

T(t)=76+C e^(-0.34t)

We know at t=0, T(0)=170

t=0→T(0)=76+C e^(-0.34(0))

170=76+C e^0

170=76+C (1)

170=76+C

Solving for C: Subtracting 76 both sides of the equation:

170-76=76+C-76

94=C

C=94

Then, the Newton's cooling model is:

T(t)=76+94 e^(-0.34t)

The temperature of the liquid after 7 hours will be:

t=7→T(7)=76+94 e^(-0.34(7))

T(7)=76+94 e^(-2.38)

T(7)=76+94 (0.092550578)

T(7)=76+8.699754332

T(7)=84.69975433

T(7)=84.7

The temperature of the liquid after 7 hours will be 84.7 °F

expression #3 is not equivalent. the others are.