Question

A pot of boiling water (212°F) is left to cool back to room temperature (70°F). How long does it take if the water cools continuously at a rate of 3% per minute?

Answer 1

To determine how long it takes for boiling water at 212°F to cool to room temperature at a continuous cooling rate of 3% per minute, we can set up an equation based on exponential decay.

Let \( T(t) \) be the temperature of the water at time \( t \) (in minutes). Given that the water cools continuously at 3% per minute, the cooling can be expressed as:

\[
T(t) = T_0 \cdot (1 - r)^t
\]

where:
- \( T_0 = 212^\circ F \) (initial temperature)
- \( r = 0.03 \) (cooling rate)
- \( T(t) = 70^\circ F \) (final temperature)

Substituting these values into the equation, we have:

\[
70 = 212 \cdot (1 - 0.03)^t
\]

This simplifies to:

\[
70 = 212 \cdot (0.97)^t
\]

Now, divide both sides by 212:

\[
\frac{70}{212} = (0.97)^t
\]

Calculating \( \frac{70}{212} \):

\[
\frac{70}{212} \approx 0.329245
\]

So we get:

\[
0.329245 = (0.97)^t
\]

To solve for \( t \), we can take the natural logarithm of both sides:

\[
\ln(0.329245) = t \cdot \ln(0.97)
\]

Now, solving for \( t \):

\[
t = \frac{\ln(0.329245)}{\ln(0.97)}
\]

Calculating the logarithms:

\[
\ln(0.329245) \approx -1.109
\]
\[
\ln(0.97) \approx -0.030459
\]

Now plug into the equation:

\[
t \approx \frac{-1.109}{-0.030459} \approx 36.4
\]

Therefore, it takes approximately **36.4 minutes** for the boiling water to cool from 212°F to 70°F.