73.7k views
4 votes
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.)

User Tlzg
by
7.3k points

1 Answer

3 votes

Answer: 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

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

User AiShiguang
by
7.8k points