Question

Нужно решить задачу по Закону Охлаждения Ньютона, максимально подробно. Задача на английском, если нужен перевод пишите An ice pack was placed in a chiller set to -4 °C. The ice pack was originally at 37 °C and it cooled to 26 °C in 16 minutes. Assuming the rate of cooling is proportional to the temperature difference between the chiller and the ice pack, calculate: a) The temperature after 60 minutes b) The time to cool within 1°C of the chiller temperature

Answer 1

Ответ:

Объяснение: кэш побьет ежже

Answer 2

To solve this problem using Newton's Law of Cooling, we can use the following formula:

$T(t) = T_{\text{chiller}} + (T_{\text{initial}} - T_{\text{chiller}})e^{-kt}$

Where:

• $T(t)$ is the temperature of the object (in this case, the ice pack) at time $t$.
• $T_{\text{chiller}}$ is the temperature of the chiller, which is -4°C in this case.
• $T_{\text{initial}}$ is the initial temperature of the ice pack, which is 37°C.
• $k$ is the cooling constant that we need to find.
• $t$ is the time in minutes.

Let's calculate $k$ first using the information given: The ice pack cools from 37°C to 26°C in 16 minutes, so we can use this data point to find $k$:

$26 = -4 + (37 - (-4))e^{-k \cdot 16}$

First, simplify the equation:

$26 = -4 + 41e^{-16k}$

Next, isolate the exponential term:

$41e^{-16k} = 26 + 4$

$41e^{-16k} = 30$

Now, divide both sides by 41:

$e^{-16k} = \frac{30}{41}$

To solve for $k$, take the natural logarithm (ln) of both sides:

$-16k = \ln\left(\frac{30}{41}\right)$

Now, solve for $k$:

$k = -\frac{1}{16}\ln\left(\frac{30}{41}\right)$

Calculate $k$ using a calculator:

$k \approx 0.0202$

Now that we have found $k$, we can use it to answer the two parts of the question:

a) The temperature after 60 minutes ($T(60)$):

$T(60) = -4 + (37 - (-4))e^{-0.0202 \cdot 60}$

Calculate $T(60)$:

$T(60) \approx -4 + 41e^{-1.212} \approx -4 + 41(0.2978) \approx -4 + 12.19 \approx 8.19°C$

So, after 60 minutes, the temperature of the ice pack will be approximately 8.19°C.

b) The time to cool within 1°C of the chiller temperature ($T(t) = -3°C$):

We need to find $t$ when $T(t) = -3$:

$-3 = -4 + (37 - (-4))e^{-0.0202t}$

Simplify the equation:

$37e^{-0.0202t} = 1$

Now, divide both sides by 37:

$e^{-0.0202t} = \frac{1}{37}$

Take the natural logarithm (ln) of both sides to solve for $t$:

$-0.0202t = \ln\left(\frac{1}{37}\right)$

Now, solve for $t$:

$t = -\frac{1}{0.0202}\ln\left(\frac{1}{37}\right)$

Calculate $t$ using a calculator:

$t \approx 134.77$

So, it will take approximately 134.77 minutes (or about 2 hours and 14 minutes) for the ice pack to cool within 1°C of the chiller temperature.