Calculate the total entropy change of the steam

[a_narain]

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?

 

[Andrew Mason]

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.

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

$$\Delta S_{water} = \int_{373}^{273} mcdT/T$$

AM

 

[nlightner]

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.