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
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
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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.
$1025 \approx 1000 = 10^3$.
$V \approx 9.75 \times 10^{13}\text{ m}^3$.
$h \approx \frac{9.75 \times 10^{13}}{3.6 \times 10^{14}} = \frac{9.75}{36} \approx 0.27\text{ m}$, which points directly to 0.25 m.
Updated On: Jun 16, 2026
- 0.25
- 0.45
- 0.15
- 0.50
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The problem asks for the vertical rise in the sea level over a span of 75 years, given the rate of ice melting on land, the density of seawater, and the total surface area of the oceans.
Step 2: Key Formula or Approach:
- Find the total mass of melted ice in 75 years.
- Convert this mass to volume using the density of seawater:
\[ V = \frac{M}{\rho} \]
- Find the rise in height $h$ by dividing the volume by the ocean surface area $A$:
\[ h = \frac{V}{A} \]
Step 3: Detailed Explanation:
• The rate of ice melting is $R = 1.3 \times 10^{15}\text{ kg/year}$.
• Over a period of $t = 75\text{ years}$, the total mass of water added to the oceans is:
\[ M_{\text{total}} = R \times t = (1.3 \times 10^{15}\text{ kg/year}) \times 75\text{ years} \]
\[ M_{\text{total}} = 9.75 \times 10^{16}\text{ kg} \]
• Now we convert this mass of water into equivalent volume $V$ using the density of seawater $\rho = 1025\text{ kg/m}^3$:
\[ V = \frac{M_{\text{total}}}{\rho} = \frac{9.75 \times 10^{16}\text{ kg}}{1025\text{ kg/m}^3} \]
\[ V \approx 9.512 \times 10^{13}\text{ m}^3 \]
• The rise in sea level $h$ is the volume divided by the surface area of the oceans $A = 3.6 \times 10^{14}\text{ m}^2$:
\[ h = \frac{V}{A} = \frac{9.512 \times 10^{13}\text{ m}^3}{3.6 \times 10^{14}\text{ m}^2} \]
\[ h \approx 0.264\text{ m} \]
• Looking at the options, $0.264\text{ m}$ is closest to $0.25\text{ m}$.
Step 4: Final Answer:
The estimated average rise in sea level is closest to 0.25 m.
The problem asks for the vertical rise in the sea level over a span of 75 years, given the rate of ice melting on land, the density of seawater, and the total surface area of the oceans.
Step 2: Key Formula or Approach:
- Find the total mass of melted ice in 75 years.
- Convert this mass to volume using the density of seawater:
\[ V = \frac{M}{\rho} \]
- Find the rise in height $h$ by dividing the volume by the ocean surface area $A$:
\[ h = \frac{V}{A} \]
Step 3: Detailed Explanation:
• The rate of ice melting is $R = 1.3 \times 10^{15}\text{ kg/year}$.
• Over a period of $t = 75\text{ years}$, the total mass of water added to the oceans is:
\[ M_{\text{total}} = R \times t = (1.3 \times 10^{15}\text{ kg/year}) \times 75\text{ years} \]
\[ M_{\text{total}} = 9.75 \times 10^{16}\text{ kg} \]
• Now we convert this mass of water into equivalent volume $V$ using the density of seawater $\rho = 1025\text{ kg/m}^3$:
\[ V = \frac{M_{\text{total}}}{\rho} = \frac{9.75 \times 10^{16}\text{ kg}}{1025\text{ kg/m}^3} \]
\[ V \approx 9.512 \times 10^{13}\text{ m}^3 \]
• The rise in sea level $h$ is the volume divided by the surface area of the oceans $A = 3.6 \times 10^{14}\text{ m}^2$:
\[ h = \frac{V}{A} = \frac{9.512 \times 10^{13}\text{ m}^3}{3.6 \times 10^{14}\text{ m}^2} \]
\[ h \approx 0.264\text{ m} \]
• Looking at the options, $0.264\text{ m}$ is closest to $0.25\text{ m}$.
Step 4: Final Answer:
The estimated average rise in sea level is closest to 0.25 m.
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