Question

problem set #7 begin date: 10/24/2025 12:01:00 am due date: 10/31/2025 11:59:00 pm end date: 11/2/2025 11:59:00 pm problem 9: (7% of assignment value) estimate the amount of water there could be in a global (planet - wide) region of subsurface permafrost on mars. that is, assume the pe surface of the planet. for these calculations, assume a permafrost thickness of 4.3 km and a concentration of ice in the permafrost of 10 - part (a) what is the surface area a of mars, in m², given that the planets radius r is 3395 km? a = 1.44 10¹⁴ a = 1.440 × 10¹⁴ m² correct! - part (b) what is the volume v of the permafrost zone, in m³? v = 6.192 10¹⁷ v = 6.192 × 10¹⁷ m³ correct! part (c) if the ice is 10% by volume, what is the total mass m of ice in this layer, in kilograms? the density of water is 1000 kg/m³. m = kg hints: deduction per hint. hints remaining: 1 feedback: 1% deduction per feedback

Explanation:

Step1: Find the volume of ice

The volume of the permafrost zone is \(V = 6.192\times10^{17}\text{ m}^3\). Since ice is \(10\%\) (or \(0.1\) in decimal) by volume, the volume of ice \(V_{\text{ice}}=0.1\times V\).

$$V_{\text{ice}} = 0.1\times6.192\times 10^{17}\text{ m}^3=6.192\times 10^{16}\text{ m}^3$$

Step2: Calculate the mass of ice

We know the formula \(
ho=\frac{M}{V}\), where \(
ho\) is density, \(M\) is mass, and \(V\) is volume. Rearranging for \(M\) gives \(M=
ho V\). Given \(
ho = 1000\text{ kg/m}^3=1\times10^{3}\text{ kg/m}^3\) and \(V = V_{\text{ice}}=6.192\times 10^{16}\text{ m}^3\).

$$M=(1\times 10^{3}\text{ kg/m}^3)\times(6.192\times 10^{16}\text{ m}^3)$$

Using the rule of exponents \(a^m\times a^n=a^{m + n}\), we have \(M = 6.192\times10^{3 + 16}\text{ kg}\)

$$M=6.192\times 10^{19}\text{ kg}$$

Answer:

\(6.192\times 10^{19}\text{ kg}\)