Change of Entropy Problem

In summary, the melting of the 4.50 kg block of ice at 0 C into the ocean causes a change in entropy for the second process, which can be calculated using the equation dS = ∫dQ/T, where S is the entropy, Q is the heat energy, and T is the absolute temperature. The temperature of the ocean may slightly decrease, but the energy for melting the ice and raising its temperature comes from the slightly warmer ocean.
  • #1
jfislove
3
0

Homework Statement



A 4.50 kg block of ice at 0 C falls into the ocean and melts. The average temperature of the ocean is 3.50 C, including all the deep water. By how much does the melting of this ice change the entropy of the world?

Homework Equations



Not sure if I should be using any equations as such, I've just been multiplying known values to get the energy, temperature and mass of the water and ice.

The Attempt at a Solution



I googled the mass of all water in the Earth's oceans, and got a figure of 7.61x10^23 kg, so I multiplied the specific heat capacity of water (4190 J/kg.K) by this mass to try and get the entropy of all the water, so I got 3.19x10^23 J/K.

I then got the entropy of the ice block by multiplying the latent heat of fusion of ice (3.34x10^5 J/kg) by the mass of the block and dividing it by the absolute temperature of the ice (273.15 K), giving what I thought was the entropy of the block to be 5502.47 J/K.

Obviously, because the two numbers are on completely different scales, adding/subtracting them makes almost no difference to the answer, so I've really no idea how to handle the question from here on!

All help is appreciated!
 
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  • #2
Tw0 things happen to the water in the ice. First the ice melts, and then the water in the ice that melted changes temperature.

How is the entropy change for the second process calculated?

Does the temperature of the ocean change? From where does the energy come that melts the ice and raises its temperature?
 
  • #3
"How is the entropy change for the second process calculated?"

Not sure how to calculate ANY entropy process, the textbook I have here (Sears and Zemansky's University Physics) gives me an equation dS = ∫dQ/T, where S is the entropy, Q is the heat energy and T is the absolute temperature, but I'm not sure how to apply that to what I have, I guess the ice block's temperature changes to the 3.5 C, but then where do I get the dQ value?

"Does the temperature of the ocean change? From where does the energy come that melts the ice and raises its temperature?"

I assume the temperature of the ocean drops SLIGHTLY, but I can't imagine it drops by any noticeable amount? Surely the energy comes from the slightly warmer ocean?
 
  • #4
jfislove said:
but then where do I get the dQ value?

From the definition of specific heat.
 
  • #5




As a scientist, it is important to approach problems like this using the appropriate equations and concepts. In this case, the concept of entropy, which is a measure of the disorder or randomness of a system, is relevant. The change in entropy of the world can be calculated by considering the entropy changes of the ice block and the ocean separately.

First, let's consider the entropy change of the ice block. As it melts, the ice changes from a solid to a liquid, which increases the disorder of its molecules. This change in entropy can be calculated using the equation ΔS = Q/T, where ΔS is the change in entropy, Q is the heat added to the system, and T is the temperature at which the heat is added.

In this case, the heat added to the ice block is the latent heat of fusion, which is 3.34x10^5 J/kg. The temperature at which this heat is added is 0°C or 273.15 K. Therefore, the change in entropy of the ice block can be calculated as:

ΔS = (3.34x10^5 J/kg) / (273.15 K) = 1223.4 J/K

Next, let's consider the entropy change of the ocean. The ocean is already at a higher temperature (3.5°C or 276.65 K) than the ice block, so the heat added to it will cause a smaller change in entropy. We can use the same equation as before to calculate this change:

ΔS = (Q/T) = (3.34x10^5 J/kg) / (276.65 K) = 1207.4 J/K

So, the total change in entropy of the world can be calculated by adding the two values together:

ΔS = 1223.4 J/K + 1207.4 J/K = 2430.8 J/K

This means that the melting of the 4.50 kg block of ice will result in a change of entropy of approximately 2430.8 J/K for the world. It is important to note that this is a very small change compared to the total entropy of the world, which is estimated to be on the order of 10^44 J/K. However, even small changes in entropy can have significant effects on the behavior of systems, which is why it is important to consider them in scientific analyses.
 

1. What is the definition of entropy?

Entropy is a measure of the disorder or randomness in a system. It is often described as the amount of energy that is not available for work in a system.

2. How does entropy relate to the concept of change?

In thermodynamics, the change in entropy is a measure of the amount of energy that is converted from a usable form to a less usable form during a process. This change is a result of the increase in disorder or randomness in the system.

3. What factors can cause a change in entropy?

A change in entropy can be caused by various factors such as changes in temperature, pressure, and chemical composition of a system. It can also be influenced by the amount of energy and the rate at which it is transferred or transformed within the system.

4. Can entropy be reversed or decreased?

According to the Second Law of Thermodynamics, the total entropy of a closed system will always increase over time. However, it is possible to decrease the entropy in one part of a system by increasing it in another part. This can be achieved through energy input and manipulation of the system.

5. How is the change of entropy problem relevant in scientific research?

The concept of entropy and its change is relevant in various fields of science, such as chemistry, physics, and biology. It helps in understanding the behavior and transformations of matter and energy in natural systems. It is also a crucial factor in determining the efficiency and feasibility of processes and reactions in different scientific experiments and industries.

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