The table shows the temperature of an amount of water set on a stove to boil, recorded every half minute.

Waiting for Water to Boil

\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline Time [tex]$(min)$[/tex] & 0 & 0.5 & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 & 3.5 & 4 & 4.5 \\
\hline Temp. [tex]$\left(^{\circ}C\right)$[/tex] & 75 & 79 & 83 & 86 & 89 & 91 & 93 & 94 & 95 & 95.5 \\
\hline
\end{tabular}

According to the line of best fit, at what time will the temperature reach [tex]$100^{\circ}C$[/tex], the boiling point of water?

A. 5 minutes

B. 5.5 minutes

C. 6 minutes

D. 6.5 minutes



Answer :

To find out when the temperature of the water will reach [tex]\(100^\circ C\)[/tex], we need to use the concept of the line of best fit. This line best represents the trend in the given data points and allows us to predict future values.

Let's break down the solution step-by-step:

1. Plotting the Data:
We have the given data points:
- Time (in minutes): [tex]\([0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4, 4.5]\)[/tex]
- Temperature (in [tex]\(^\circ C\)[/tex]): [tex]\([75, 79, 83, 86, 89, 91, 93, 94, 95, 95.5]\)[/tex]

2. Finding the Line of Best Fit:
The line of best fit can be represented by the equation of a straight line:
[tex]\[ \text{Temperature} = \text{slope} \times \text{time} + \text{intercept} \][/tex]

From our calculations, the slope ([tex]\(m\)[/tex]) and the intercept ([tex]\(b\)[/tex]) are found to be:
- Slope ([tex]\(m\)[/tex]): [tex]\(4.539393939393942\)[/tex]
- Intercept ([tex]\(b\)[/tex]): [tex]\(77.83636363636359\)[/tex]

So, the equation of the line of best fit is:
[tex]\[ \text{Temperature} = 4.539393939393942 \times \text{time} + 77.83636363636359 \][/tex]

3. Solving for Time when Temperature is 100°C:
We want to find the time ([tex]\(t\)[/tex]) when the temperature ([tex]\(T\)[/tex]) is [tex]\(100^\circ C\)[/tex]:
[tex]\[ 100 = 4.539393939393942 \times t + 77.83636363636359 \][/tex]

4. Isolating the Variable [tex]\(t\)[/tex]:
Subtract the intercept from both sides:
[tex]\[ 100 - 77.83636363636359 = 4.539393939393942 \times t \][/tex]
[tex]\[ 22.16363636363641 = 4.539393939393942 \times t \][/tex]

5. Solving for [tex]\(t\)[/tex]:
Divide both sides by the slope:
[tex]\[ t = \frac{22.16363636363641}{4.539393939393942} \][/tex]

6. Calculating the Time:
[tex]\[ t \approx 4.882510013351143 \][/tex]

Thus, according to the line of best fit, the temperature of the water will reach [tex]\(100^\circ C\)[/tex] in approximately [tex]\(4.88\)[/tex] minutes. This is very close to 5 minutes, but not exceeding it. Considering conventional rounding, the closest given option to [tex]\(4.88\)[/tex] is [tex]\(5\)[/tex].

Therefore, the time at which the temperature will reach [tex]\(100^\circ C\)[/tex] is:
[tex]\[ \boxed{5} \][/tex]