# Calculate the total entropy change of the steam

1. May 3, 2009

### a_narain

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Entropy = integral of dQ/T
Q = mL + mc(delta T)

3. 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?

2. May 3, 2009

### Andrew Mason

Re: Entropy Problem

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