Question

Ice on land mass melting into ocean, due to global warming, is estimated to be around $1.3 \times 10^{15}\text{ kg}$ per year. The density of sea water is $1025\text{ kg}\cdot\text{m}^{-3}$. Assuming that the melting rate of ice remains constant and the effective surface area covered by the oceans is $3.6 \times 10^{14}\text{ m}^2$, the estimated average rise in sea level (in m) in the next 75 years will be closest to

• A. 0.25
• B. 0.45
• C. 0.15
• D. 0.50

Hint:

Using quick approximation:
$1025 \approx 1000 = 10^3$.
$V \approx 9.75 \times 10^{13}\text{ m}^3$.
$h \approx \frac{9.75 \times 10^{13}}{3.6 \times 10^{14}} = \frac{9.75}{36} \approx 0.27\text{ m}$, which points directly to 0.25 m.

Answer 1

The Correct Option is A

Step 1: See the plan.
Melted ice adds water to the ocean. We find the total mass added over 75 years, turn it into a volume, and spread that volume over the ocean surface to get the rise in height.

Step 2: Find the total mass added.
Each year about $1.3 \times 10^{15}$ kg melts, so over 75 years \[ M = 1.3 \times 10^{15} \times 75 \approx 9.75 \times 10^{16}\ \text{kg}. \]

Step 3: Convert mass to volume.
Using sea water density $\rho = 1025$ kg per cubic metre, \[ V = \frac{M}{\rho} = \frac{9.75 \times 10^{16}}{1025} \approx 9.5 \times 10^{13}\ \text{m}^3. \]

Step 4: Recall the spread-out idea.
If this volume sits as a thin layer over the oceans, then volume equals area times height, so height $= V / A$.

Step 5: Divide by the ocean area.
With $A = 3.6 \times 10^{14}$ square metres, \[ h = \frac{9.5 \times 10^{13}}{3.6 \times 10^{14}} \approx 0.26\ \text{m}. \]

Step 6: Round to the nearest option.
The rise is closest to a quarter of a metre. \[ \boxed{h \approx 0.25\ \text{m}} \]