Calculate the total entropy change of the steam

In summary, the problem involves a system with 50kg of steam at 100C and a heat reservoir at 0C. A reversible heat engine is used to gradually condense and cool the steam until it reaches 0C. The task is to calculate the total entropy change of the steam and subsequent water. The equation used is the integral of dQ/T, where Q is equal to the sum of heat of vaporization and heat flow as the water cools. The correct answer is -368 kJ, which can be calculated by evaluating the integral between the temperatures in Kelvin. The mistake in the attempt at solving the problem was taking the derivative instead of evaluating the integral.
  • #1
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Homework Statement


You have 50kg of steam at 100C, but no heat source to maintain it in that condition. You also have a heat resevoir at 0C. Suppose you operate a reversible heat engine with this system, so that the steam gradually condenses and cools until it reaches 0C. Calculate the total entropy change of the steam and subsequent water.


Homework Equations


Entropy = integral of dQ/T
Q = mL + mc(delta T)


The Attempt at a Solution


I set Q= ml+mc(delta T) and took the derivative to find dQ in terms of T. I then evaluated the integral between the two temperatures in kelvin, and got -70 kJ. However, the answer is -368 kJ. What am I doing wrong?
 
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  • #2


a_narain said:

Homework Statement


You have 50kg of steam at 100C, but no heat source to maintain it in that condition. You also have a heat resevoir at 0C. Suppose you operate a reversible heat engine with this system, so that the steam gradually condenses and cools until it reaches 0C. Calculate the total entropy change of the steam and subsequent water.

Homework Equations


Entropy = integral of dQ/T
Q = mL + mc(delta T)

The Attempt at a Solution


I set Q= ml+mc(delta T) and took the derivative to find dQ in terms of T. I then evaluated the integral between the two temperatures in kelvin, and got -70 kJ. However, the answer is -368 kJ. What am I doing wrong?
Why are you taking the derivative? There are two parts to the integral. There is the heat of vaporisation. Then there is the heat flow as the water cools to 0 C.

[tex]\Delta S_{steam} = \int dQ/T = mL/T[/tex] where T = 100C (373K)

[tex]\Delta S_{water} = \int_{373}^{273} mcdT/T[/tex]

AM
 
  • #3


First of all, it is important to note that the given problem does not specify the units for the heat capacity (c) and latent heat (L). Therefore, it is not possible to accurately calculate the total entropy change without knowing these values.

However, assuming that the units for c and L are in kJ/kg-K and kJ/kg respectively, there are a few issues with your approach.

1. The equation you used, Q = mL + mc(delta T), is only valid for a single phase system. In this case, the steam undergoes a phase change from gas to liquid, so the equation should be written as Q = mL + mLv + mc(delta T), where Lv is the latent heat of vaporization.

2. Taking the derivative of this equation to find dQ in terms of T is not correct. The integral of dQ/T should be evaluated directly, without taking the derivative.

3. The integral should be evaluated from 373 K (100C) to 273 K (0C), not in kelvin, but in degrees Celsius. This is because the heat capacity and latent heat are given in terms of temperature in degrees Celsius.

4. The correct integral to evaluate would be: ∫(mL + mLv + mc(delta T))/T dT from 100C to 0C.

5. Finally, make sure to convert the final entropy change from kJ/K to kJ. This can be done by multiplying the result by the temperature difference (100C - 0C = 100K).

With these corrections, the total entropy change of the steam and subsequent water can be accurately calculated.
 

1. What is entropy change?

Entropy change is a measure of the disorder or randomness of a system. In the context of steam, it refers to the change in the distribution or arrangement of molecules as it undergoes a physical or chemical process.

2. How is entropy change calculated?

The total entropy change of steam can be calculated by considering the entropy changes at each stage of the process, using the formula ΔS = Q/T, where ΔS is the change in entropy, Q is the heat transferred, and T is the temperature in Kelvin.

3. What factors affect the entropy change of steam?

The entropy change of steam is affected by the temperature, pressure, and volume of the system. Changes in these parameters can lead to changes in the distribution and arrangement of molecules, resulting in a change in entropy.

4. What is the significance of calculating the total entropy change of steam?

Calculating the total entropy change of steam is important in understanding the efficiency of a process and predicting the direction of spontaneous reactions. It also allows for the optimization of systems to minimize energy loss.

5. Can entropy change be negative?

Yes, entropy change can be negative. This indicates that the system has become more ordered or less random. However, in accordance with the second law of thermodynamics, the total entropy of the universe will always increase.

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