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Chapter 9: Problem 26

If the ocean absorbs an additional \(3 \mathrm{~W} / \mathrm{m}^{2}\) of forcing, \({ }^{86}\) how long would it take to heat the \(\sim 4 \mathrm{~km}\) deep column of water directly under a particular square meter (thus \(4 \times 10^{6} \mathrm{~kg}\) ) by \(1^{\circ} \mathrm{C} ?^{87}\)

Short Answer

Expert verified
Answer: It would take approximately 64,444.4 days for the 4 km deep column of water to be heated by 1°C.

Step by step solution

01

Write down known variables

Heat input (P): \(3\, \mathrm{W/m}^2\) Mass of water column (m): \(4 × 10^6\, \mathrm{kg}\) Depth of water column: \(4\, \mathrm{km}\) Temperature change (ΔT): \(1^{\circ}\mathrm{C}\) Specific heat capacity of water (c): \(4.18 × 10^3\,\mathrm{J/\! kg\cdot K}\)
02

Calculate the heat required to raise the temperature of the water column by \(1^{\circ}\mathrm{C}\)

We will use the heat formula: \(Q = mc\Delta T\) \(Q = (4 \times 10^6\, \mathrm{kg}) \cdot (4.18 × 10^3\,\mathrm{J/\! kg\cdot K}) \cdot (1^{\circ}\mathrm{C})\)
03

Simplify the expression to find the value of Q

\(Q = 4 \times 10^6\, \mathrm{kg} \cdot 4.18 × 10^3\,\mathrm{J/\! kg\cdot K} \cdot 1^{\circ}\mathrm{C}\) \(Q = 16.72 \times 10^9\, \mathrm{J}\)
04

Calculate the time required to heat the water column by the given heat input

We will use the formula for time: \(t = Q / P\) \(t = (16.72 \times 10^9\, \mathrm{J}) / (3\, \mathrm{W/m}^2)\)
05

Simplify the expression to find the value of t

\(t = (16.72 \times 10^9\, \mathrm{J}) / (3\, \mathrm{W/m}^2)\) \(t = 5.57 \times 10^9\, \mathrm{s}\) (in seconds)
06

Convert the time to a more convenient unit of measurement

From seconds to days: \(t = (5.57 \times 10^9\, \mathrm{s}) \cdot (1\, \mathrm{day} / 86400\, \mathrm{s})\) \(t \approx 64444.4\,\mathrm{days}\)
07

Write the conclusion

It would take approximately \(64444.4\,\mathrm{days}\) to heat the \(4\,\mathrm{km}\) deep column of water directly under a particular square meter by \(1^{\circ}\mathrm{C}\).

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