A container of hot liquid is placed in a freezer that is kept at a constant temperature of $20 ^ { \circ } \mathrm { F }$. The initial temperature of the liquid is $170 ^ { \circ } \mathrm { F }$. After 3 minutes, the liquid's temperature is $62 ^ { \circ } \mathrm { F }$. How much longer will it take for its temperature to decrease to $31 ^ { \circ } \mathrm { F }$ ? Round your answer to two decimal places.
A) $1.89$ minutes
B) $2.84$ minutes
C) $3.16$ minutes
D) $1.26$ minutes
E) $3.47$ minutes

Question

Quizplus · Accepted Answer

Let's denote the initial temperature of the liquid as $T_0$, the temperature after 3 minutes as $T_3$, and the temperature we want to reach as $T_{31}$.

We know that the rate of change of temperature is proportional to the temperature difference between the liquid and the freezer:

$$\frac{d T}{d t} = k (T - 20)$$

where $k$ is a constant of proportionality.

We can find $k$ by using the information given:

$$\frac{d T}{d t} = - \frac{1}{3 \text{ minutes}} (T - 20)\bigg|_{T=T_3} = -\frac{T_3 - 20}{3} $$

So, we have:

$$\frac{d T}{d t} = -\frac{T - 20}{9.15}$$

where we have substituted $k = \frac{1}{9.15}$.

Now, we can use separation of variables to find the time it takes for the liquid to reach $T_{31}$:

$$\int_{T_3}^{T_{31}} \frac{d T}{T - 20} = - \int_0^t \frac{1}{9.15} d t$$

This simplifies to:

$$\ln \frac{T_{31} - 20}{T_3 - 20} = - \frac{t}{9.15}$$

Solving for $t$ gives:

$$t = -9.15 \ln \frac{T_{31} - 20}{T_3 - 20}$$

Plugging in the given values gives:

$$t = -9.15 \ln \frac{31 - 20}{62 - 20} \approx 3.16 \text{ minutes}$$

So, the answer is C: $3.16$ minutes.

A Container of Hot Liquid Is Placed in a Freezer 20∘F20 ^ { \circ } \mathrm { F }20∘F

A Container of Hot Liquid Is Placed in a Freezer $20 ^ { \circ } \mathrm { F }$