Question

Scientists believe that early in its evolution, the Moon was covered by a magma ocean with a depth of \(50 \mathrm{~km}\). Assuming that the magma was at its melt temperature of \(1500 \mathrm{~K}\) and that the surface of the Moon was maintained at \(500 \mathrm{~K}\), how long did it take for the magma ocean to solidify if it was cooled from the surface? Take \(L=320 \mathrm{~kJ} \mathrm{~kg}^{-1}, \kappa=1 \mathrm{~mm}^{2}\) \(\mathrm{s}^{-1}\), and \(c=1 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\).

Short answer

The magma ocean took approximately 124 million years to solidify.

Step-by-step solution

Identify the Formula

To determine how long it took for the magma ocean to solidify, we can use the formula for solidification time of a layer with constant surface cooling, which is given by: \[ t = \frac{d^2}{2\kappa} \left( \frac{c \Delta T}{L} \right) \] where \(d\) is the depth of the magma ocean, \(\kappa\) is the thermal diffusivity, \(c\) is the specific heat capacity, \(L\) is the latent heat of fusion, and \(\Delta T\) is the temperature difference.

Assign Known Values

From the problem, we have:- Depth of the magma ocean, \(d = 50 \text{ km} = 50,000 \text{ m}\) (conversion from kilometers to meters).- Melt temperature, \(T_m = 1500 \text{ K}\).- Surface temperature, \(T_s = 500 \text{ K}\).- \(\kappa = 1 \text{ mm}^2/\text{s} = 1 \times 10^{-6} \text{ m}^2/\text{s}\) (conversion from square millimeters to square meters).- Specific heat capacity, \(c = 1 \text{ kJ}/\text{kg K} = 1,000 \text{ J}/\text{kg K}\) (conversion from kilojoules to joules).- Latent heat, \(L = 320 \text{ kJ}/\text{kg} = 320,000 \text{ J}/\text{kg}\) (conversion from kilojoules to joules).

Calculate Temperature Difference

Calculate the temperature difference \(\Delta T\) between the magma and the surface of the Moon: \[ \Delta T = T_m - T_s = 1500 \text{ K} - 500 \text{ K} = 1000 \text{ K} \]

Insert Values into the Formula and Calculate

Insert the known values into the formula to solve for \(t\):\[ t = \frac{(50,000)^2}{2 \times (1 \times 10^{-6})} \left( \frac{1,000 \times 1000}{320,000} \right) \] First, calculate \(\frac{c \Delta T}{L}\):\[ \frac{1,000 \times 1000}{320,000} = 3.125 \]Then calculate the time \(t\):\[ t = \frac{2,500,000,000}{2 \times 10^{-6}} \times 3.125 = 3.125 \times 1,250,000,000,000 \]\[ t \approx 3.90625 \times 10^{12} \text{ seconds} \]

Convert Time to Years

Convert time from seconds to years for an easier interpretation.There are \(60 \times 60 \times 24 \times 365.25 = 31,557,600\) seconds in a year (including leap year).\[ t = \frac{3.90625 \times 10^{12}}{31,557,600} \approx 123,794,004 \text{ years} \]

Key concepts

Solidification Time Calculation

To understand the solidification time of the Moon's magma ocean, we rely on precise calculations. Imagine the Moon's surface gradually losing heat, causing the underlying magma to cool and solidify over time. The formula for the solidification time of a layer under constant surface cooling is given by:\[ t = \frac{d^2}{2\kappa} \left( \frac{c \Delta T}{L} \right) \]This formula helps us calculate how long it took for the magma to change from a liquid to a solid state. Here's what each symbol means:

• \(t\): Time taken for solidification.
• \(d\): Depth of the magma ocean.
• \(\kappa\): Thermal diffusivity, measuring how quickly heat spreads through a material.
• \(c\): Specific heat capacity, indicating how much heat energy is required to raise the temperature.
• \(L\): Latent heat of fusion, the heat needed to convert solid to liquid without changing temperature.
• \(\Delta T\): Temperature difference between the melt and the surface.

Understanding this formula lets us appreciate the vast time span it took for the magma to solidify, calculated to be roughly 123 million years.

Thermal Diffusivity

Thermal diffusivity is a measure of how quickly heat spreads through a material. It combines thermal conductivity, density, and specific heat capacity into one measurement, denoted by \(\kappa\). For the Moon's magma ocean, the thermal diffusivity value \(1 \,\text{mm}^2/\text{s}\) helps us understand the rate at which heat leaves the magma and transfers to the moon's surface. In geophysics, knowing the thermal diffusivity is crucial as it influences how long the cooling process takes.The unit of thermal diffusivity is square meters per second (\(\text{m}^2/\text{s}\)), emphasizing the area through which heat diffuses over time. In our exercise, thermal diffusivity is essential in determining how effectively the Moon's surface lost heat, contributing to the lengthy solidification period of the magma ocean.

Latent Heat of Fusion

Latent heat of fusion \(L\) refers to the energy required to change a substance from solid to liquid at its melting point without changing its temperature. For the Moon's magma ocean, \(L = 320 \text{ kJ}/\text{kg}\) is the latent heat value. Understanding this energy is vital for lunar studies, as it directly relates to how much energy was needed for the magma to remain liquid during the moon's early days.The concept is crucial in this context because as the magma cooled, the energy stored as latent heat was released as it solidified, with this energy influencing the time it took for the thick layer of magma to transform into rock. The larger the latent heat, the more heat energy is involved in maintaining the liquid to solid phase change of the magma, influencing the cooling rate and time.

Temperature Difference in Geophysics

In geophysics, temperature difference (\(\Delta T\)) plays a pivotal role in understanding geological processes. It refers to the difference in temperature between two points—in this exercise, between the magma ocean's melt temperature (\(1500 \,\text{K}\)) and the Moon's surface temperature (\(500 \,\text{K}\)). This difference, quantified as \(1000 \,\text{K}\), drives the solidification process forward.Why is this important? The temperature difference influences the rate at which heat flows from the magma to its surroundings. The larger the \(\Delta T\), the faster the potential cooling rate, as more heat will transfer from the magma to the lunar surface. This significant gradient indicates a substantial energy transfer required to cool and solidify the vast ocean of molten rock, ultimately affecting the timescale required for solidification. Understanding this temperature difference provides vital insights into the cooling and geological evolution of celestial bodies like the Moon.