A container of hot liquid is placed in a freezer that is kept at a constant temperature of .The initial temperature of the liquid is .After 5 minutes,the liquid's temperature is .How much longer will it take for its temperature to decrease to ? Round your answer to two decimal places.
A)5.29 minutes
B)7.94 minutes
C)8.82 minutes
D)3.53 minutes
E)9.70 minutes

Question

A container of hot liquid is placed in a freezer that is kept at a constant temperature of  .The initial temperature of the liquid is  .After 5 minutes,the liquid's temperature is  .How much longer will it take for its temperature to decrease to  ? Round your answer to two decimal places. ​
A)5.29 minutes 
B)7.94 minutes 
C)8.82 minutes 
D)3.53 minutes 
E)9.70 minutes

Quizplus · Accepted Answer

We can use the formula for Newton's law of cooling to solve this problem:
T(t) = (T0 - Tf)e^(-kt) + Tf
where
T(t) is the temperature of the liquid after time t,
T0 is the initial temperature of the liquid,
Tf is the temperature of the freezer,
k is a constant that depends on the properties of the liquid and the container,
and e is the base of the natural logarithm.

We are given that T0 = 61, Tf = -16, and T(5) = 26. We want to find the time t such that T(t) = -27.

Plugging in the known values, we get:
26 = (61 - (-16))e^(-5k) - 16
42 = 77e^(-5k)
e^(-5k) = 42/77
-5k = ln(42/77)
k = -ln(42/77)/5 ≈ 0.162

Now we can plug in T0 = 61, Tf = -16, and T(t) = -27, and solve for t:
-27 = (61 - (-16))e^(-0.162t) - 16
-72 = 77e^(-0.162t)
e^(-0.162t) = -72/77
-0.162t = ln(-72/77)
t = -ln(-72/77)/0.162 ≈ 8.82

Therefore, it will take approximately 8.82 minutes for the liquid's temperature to decrease to -27°C. The correct answer is (C).

A Container of Hot Liquid Is Placed in a Freezer