To solve this problem, we can use Newton's Law of Cooling, which is given by:

$T(t) = T_{\text{env}} + (T_0 - T_{\text{env}}) e^{-kt}$

Where:

• $T(t)$ is the temperature of the object at time $t$.
• $T_{\text{env}}$ is the ambient temperature (0°C in this case).
• $T_0$ is the initial temperature of the object (21°C).
• $k$ is the cooling constant.
• $t$ is the time elapsed.

Step 1: Determine the cooling constant $k$

First, we need to find $k$ using the information given:

At $t = 1$ minute, the temperature drops to 10°C:

$10 = 0 + (21 - 0) e^{-k \cdot 1}$

$10 = 21 e^{-k}$

$\frac{10}{21} = e^{-k}$

Taking the natural logarithm on both sides:

$\ln \left(\frac{10}{21}\right) = -k$

$k = -\ln \left(\frac{10}{21}\right)$

Step 2: Determine the time $t$ when the temperature is 5°C

Now, we need to find the time $t$ when the temperature is 5°C:

$5 = 0 + (21 - 0) e^{-kt}$

$\frac{5}{21} = e^{-kt}$

Taking the natural logarithm on both sides:

$\ln \left(\frac{5}{21}\right) = -kt$

Substituting the value of $k$ we found earlier:

$\ln \left(\frac{5}{21}\right) = -\left(-\ln \left(\frac{10}{21}\right)\right)t$

$t = \frac{\ln \left(\frac{5}{21}\right)}{-\ln \left(\frac{10}{21}\right)}$

Step 3: Calculate the value of $t$

Let's compute this value.

$t = \frac{\ln \left(\frac{5}{21}\right)}{-\ln \left(\frac{10}{21}\right)}$

$t = \frac{\ln \left(\frac{5}{21}\right)}{\ln \left(\frac{21}{10}\right)}$

Calculating the logarithms:

$t \approx \frac{\ln(0.2381)}{\ln(2.1)} \approx \frac{-1.433}{0.741}$

$t \approx 1.934 \text{ minutes}$

Final Answer

It will take approximately 1.934 minutes after removal for the thermometer reading to be 5°C.

Would you like further details or have any questions?

Here are 5 related questions for practice:

1. What is Newton's Law of Cooling and when is it applied?
2. How do you determine the cooling constant $k$ from experimental data?
3. How does changing the ambient temperature affect the cooling time?
4. What would happen if the thermometer was taken to an environment with a temperature below 0°C?
5. Can Newton's Law of Cooling be used to estimate the time of death in forensic science?

Tip: Always take the natural logarithm of both sides when solving exponential equations for time, as it helps isolate the variable of interest.