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Answer :
Below is a detailed step‐by‐step solution.
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■ Part 3.1: The Cooling of the Chokka
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The temperature (in °C) at any time (in minutes) is given by
$$
T = 18 - \left(14.5 \times \frac{\text{time in minutes}}{60}\right).
$$
▪︎ 3.1.1 – Finding the Missing Values D and E
1. For the temperature at 120 minutes (denoted by D):
Substitute 120 for time:
$$
T = 18 - \left(14.5 \times \frac{120}{60}\right) = 18 - \left(14.5 \times 2\right) = 18 - 29 = -11^\circ C.
$$
Thus, $$\boxed{D = -11^\circ C}.$$
2. For the time E when the temperature is 0°C, we set $T=0$ and solve for the time:
$$
0 = 18 - \left(14.5 \times \frac{E}{60}\right).
$$
Rearrange to obtain:
$$
14.5 \times \frac{E}{60} = 18 \quad \Longrightarrow \quad E = \frac{18 \times 60}{14.5}.
$$
Calculating that gives approximately
$$
E \approx 74.48\text{ minutes}.
$$
Thus, $$\boxed{E \approx 74.48\text{ minutes}}.$$
▪︎ 3.1.3 – Verifying the Freezer's 10-Minute Cooling Claim
A crew member claims that the chokka cools at a constant rate of $2.42^\circ C$ every 10 minutes.
To check the claim, we calculate the temperature drop in 10 minutes using the formula:
$$
\text{Cooling in 10 minutes} = 14.5 \times \frac{10}{60} = \frac{14.5}{6} \approx 2.41667^\circ C.
$$
Comparing the calculated rate (approximately $2.41667^\circ C$ per 10 minutes) with the claimed rate ($2.42^\circ C$ per 10 minutes), we see that the difference is very small (around $0.00333^\circ C$). This confirms that the cooling rate is effectively constant at the stated rate.
Thus, the claim is valid.
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■ Part 3.2: Crude Oil Barrel Calculations
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For a standard barrel, we assume it holds 42 gallons and has a height of 18 inches. Recall that 1 gallon is equivalent to 231 cubic inches. Hence, the volume of the barrel in cubic inches is
$$
\text{Volume} = 42 \times 231.
$$
The barrel is a cylinder with volume given by
$$
\pi r^2 h,
$$
where $h = 18$ inches. Solving for the radius $r$, we have
$$
r = \sqrt{\frac{\text{Volume}}{\pi \times 18}}.
$$
With the calculated volume, this gives approximately
$$
r \approx 13.10\text{ inches}.
$$
Next, the full surface area (including the top and bottom) of a cylinder is
$$
\text{Surface Area} = 2\pi r^2 + 2\pi r h.
$$
Substituting $r \approx 13.10$ inches and $h = 18$ inches results in a surface area of approximately
$$
\text{Surface Area} \approx 2559.40 \text{ square inches}.
$$
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■ Part 3.3: Moisturising Gel Container Calculations
────────────────────────────
Consider a cylindrical container with a volume of 500 ml (recalling that $1\,\text{ml} = 1\,\text{cm}^3$) and an assumed radius of 4 cm.
1. 3.3.1 – Height of the Container
The volume of a cylinder is given by
$$
\text{Volume} = \pi r^2 h.
$$
Solving for the height $h$, we have
$$
h = \frac{\text{Volume}}{\pi r^2} = \frac{500}{\pi (4)^2} = \frac{500}{16\pi}.
$$
This produces
$$
h \approx 9.95\text{ cm}.
$$
2. 3.3.2 – Percentage of the Curved Surface Area of the Total Surface Area
The lateral (curved) surface area (CSA) of a cylinder is
$$
\text{CSA} = 2\pi r h,
$$
and the total surface area (TSA) is
$$
\text{TSA} = 2\pi r^2 + 2\pi r h.
$$
Thus, the percentage of the curved surface area is
$$
\text{Percentage} = \frac{\text{CSA}}{\text{TSA}} \times 100 = \frac{2\pi r h}{2\pi r^2 + 2\pi r h} \times 100.
$$
Substituting $r = 4\text{ cm}$ and $h \approx 9.95\text{ cm}$ yields approximately
$$
\text{Percentage} \approx 71.32\%.
$$
────────────────────────────
■ Final Answers
────────────────────────────
1. The missing values in the table are:
$$\boxed{D = -11^\circ C, \quad E \approx 74.48\text{ minutes}}.$$
2. A line graph (not shown here) should plot the temperature against time with the points:
$$ (0, 18^\circ C), \quad (E \approx 74.48\text{ minutes}, \, 0^\circ C), \quad (120\,\text{min}, -11^\circ C), \quad (180\,\text{min}, -25.5^\circ C), \quad (240\,\text{min}, -40^\circ C),$$
and a constant temperature of $-40^\circ C$ at $360$ minutes.
3. The freezer cools the chokka at a rate of about $2.41667^\circ C$ every 10 minutes, which is consistent with the claimed rate of $2.42^\circ C$ (the small discrepancy being negligible).
4. For the crude oil barrel:
The radius is approximately $$\boxed{13.10\text{ inches}},$$
and the total surface area is approximately $$\boxed{2559.40\text{ square inches}}.$$
5. For the moisturising gel container:
The height is approximately $$\boxed{9.95\text{ cm}},$$
and the curved surface area constitutes about $$\boxed{71.32\%}$$ of the total surface area.
This completes the detailed solution for all parts of the question.
────────────────────────────
■ Part 3.1: The Cooling of the Chokka
────────────────────────────
The temperature (in °C) at any time (in minutes) is given by
$$
T = 18 - \left(14.5 \times \frac{\text{time in minutes}}{60}\right).
$$
▪︎ 3.1.1 – Finding the Missing Values D and E
1. For the temperature at 120 minutes (denoted by D):
Substitute 120 for time:
$$
T = 18 - \left(14.5 \times \frac{120}{60}\right) = 18 - \left(14.5 \times 2\right) = 18 - 29 = -11^\circ C.
$$
Thus, $$\boxed{D = -11^\circ C}.$$
2. For the time E when the temperature is 0°C, we set $T=0$ and solve for the time:
$$
0 = 18 - \left(14.5 \times \frac{E}{60}\right).
$$
Rearrange to obtain:
$$
14.5 \times \frac{E}{60} = 18 \quad \Longrightarrow \quad E = \frac{18 \times 60}{14.5}.
$$
Calculating that gives approximately
$$
E \approx 74.48\text{ minutes}.
$$
Thus, $$\boxed{E \approx 74.48\text{ minutes}}.$$
▪︎ 3.1.3 – Verifying the Freezer's 10-Minute Cooling Claim
A crew member claims that the chokka cools at a constant rate of $2.42^\circ C$ every 10 minutes.
To check the claim, we calculate the temperature drop in 10 minutes using the formula:
$$
\text{Cooling in 10 minutes} = 14.5 \times \frac{10}{60} = \frac{14.5}{6} \approx 2.41667^\circ C.
$$
Comparing the calculated rate (approximately $2.41667^\circ C$ per 10 minutes) with the claimed rate ($2.42^\circ C$ per 10 minutes), we see that the difference is very small (around $0.00333^\circ C$). This confirms that the cooling rate is effectively constant at the stated rate.
Thus, the claim is valid.
────────────────────────────
■ Part 3.2: Crude Oil Barrel Calculations
────────────────────────────
For a standard barrel, we assume it holds 42 gallons and has a height of 18 inches. Recall that 1 gallon is equivalent to 231 cubic inches. Hence, the volume of the barrel in cubic inches is
$$
\text{Volume} = 42 \times 231.
$$
The barrel is a cylinder with volume given by
$$
\pi r^2 h,
$$
where $h = 18$ inches. Solving for the radius $r$, we have
$$
r = \sqrt{\frac{\text{Volume}}{\pi \times 18}}.
$$
With the calculated volume, this gives approximately
$$
r \approx 13.10\text{ inches}.
$$
Next, the full surface area (including the top and bottom) of a cylinder is
$$
\text{Surface Area} = 2\pi r^2 + 2\pi r h.
$$
Substituting $r \approx 13.10$ inches and $h = 18$ inches results in a surface area of approximately
$$
\text{Surface Area} \approx 2559.40 \text{ square inches}.
$$
────────────────────────────
■ Part 3.3: Moisturising Gel Container Calculations
────────────────────────────
Consider a cylindrical container with a volume of 500 ml (recalling that $1\,\text{ml} = 1\,\text{cm}^3$) and an assumed radius of 4 cm.
1. 3.3.1 – Height of the Container
The volume of a cylinder is given by
$$
\text{Volume} = \pi r^2 h.
$$
Solving for the height $h$, we have
$$
h = \frac{\text{Volume}}{\pi r^2} = \frac{500}{\pi (4)^2} = \frac{500}{16\pi}.
$$
This produces
$$
h \approx 9.95\text{ cm}.
$$
2. 3.3.2 – Percentage of the Curved Surface Area of the Total Surface Area
The lateral (curved) surface area (CSA) of a cylinder is
$$
\text{CSA} = 2\pi r h,
$$
and the total surface area (TSA) is
$$
\text{TSA} = 2\pi r^2 + 2\pi r h.
$$
Thus, the percentage of the curved surface area is
$$
\text{Percentage} = \frac{\text{CSA}}{\text{TSA}} \times 100 = \frac{2\pi r h}{2\pi r^2 + 2\pi r h} \times 100.
$$
Substituting $r = 4\text{ cm}$ and $h \approx 9.95\text{ cm}$ yields approximately
$$
\text{Percentage} \approx 71.32\%.
$$
────────────────────────────
■ Final Answers
────────────────────────────
1. The missing values in the table are:
$$\boxed{D = -11^\circ C, \quad E \approx 74.48\text{ minutes}}.$$
2. A line graph (not shown here) should plot the temperature against time with the points:
$$ (0, 18^\circ C), \quad (E \approx 74.48\text{ minutes}, \, 0^\circ C), \quad (120\,\text{min}, -11^\circ C), \quad (180\,\text{min}, -25.5^\circ C), \quad (240\,\text{min}, -40^\circ C),$$
and a constant temperature of $-40^\circ C$ at $360$ minutes.
3. The freezer cools the chokka at a rate of about $2.41667^\circ C$ every 10 minutes, which is consistent with the claimed rate of $2.42^\circ C$ (the small discrepancy being negligible).
4. For the crude oil barrel:
The radius is approximately $$\boxed{13.10\text{ inches}},$$
and the total surface area is approximately $$\boxed{2559.40\text{ square inches}}.$$
5. For the moisturising gel container:
The height is approximately $$\boxed{9.95\text{ cm}},$$
and the curved surface area constitutes about $$\boxed{71.32\%}$$ of the total surface area.
This completes the detailed solution for all parts of the question.
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