What is the entropy change of the meteor

[kitkat2950]

A 3400kg meteor (c=450 J/KgC) at 1250 degrees C falls into lake michigan, which is at a temperature of 12 degrees C. Assuming the temperature of the lake remains constant,

A) What is the entropy change of the meteor as it reaches thermal equilibrium with Lake Michigan?
B) What is the total entropy change of the universe? (meteor and Lake Michigan system)

I was able to get the first part and found the entropy change of the meteor to be -2.564E6, but I'm having trouble with part be. I think the equation I need is Entropy of universe= entropy of meteor + entropy of lake. However, I have no idea how to find the entropy of the lake. Please help. I also know the answer is 4.1E6.

 

[Andrew Mason]

A 3400kg meteor (c=450 J/KgC) at 1250 degrees C falls into lake michigan, which is at a temperature of 12 degrees C. Assuming the temperature of the lake remains constant,

A) What is the entropy change of the meteor as it reaches thermal equilibrium with Lake Michigan?
B) What is the total entropy change of the universe? (meteor and Lake Michigan system)

I was able to get the first part and found the entropy change of the meteor to be -2.564E6, but I'm having trouble with part be. I think the equation I need is Entropy of universe= entropy of meteor + entropy of lake. However, I have no idea how to find the entropy of the lake. Please help. I also know the answer is 4.1E6.

What is the expression for change in entropy? How much heat flows from the meteor to the lake? Does the temperature of the lake change during the heat transfer from the meteor to the lake? If you answer these questions, you will easily find the entropy change in the lake.

The entropy change of the meteor is a little more difficult because the temperature of the meteor is decreasing as the heat flows out of it.

AM

 

[baba89]

A) The entropy change of the meteor as it reaches thermal equilibrium with Lake Michigan can be calculated using the formula ΔS = mCln(Tf/Ti), where m is the mass of the meteor, C is its specific heat capacity, and Tf and Ti are the final and initial temperatures, respectively. Plugging in the given values, we get:

ΔS = (3400 kg)(450 J/kg°C) ln(12°C/1250°C) = -2.564 x 10^6 J/K

B) To find the total entropy change of the universe, we need to consider the entropy change of both the meteor and the lake. The entropy change of the lake can be calculated using the same formula as above, but with the mass and specific heat capacity of the lake. However, since the temperature of the lake remains constant, the final and initial temperatures will be the same, and the natural logarithm will equal 1, resulting in an entropy change of 0 for the lake.

Therefore, the total entropy change of the universe is simply the entropy change of the meteor, which we calculated in part A:

ΔS = -2.564 x 10^6 J/K

This means that the total entropy of the universe decreases as the meteor reaches thermal equilibrium with the lake, indicating a decrease in disorder or randomness. However, this decrease in entropy is only temporary and will eventually increase again due to natural processes.