Entropy change when converting water to steam

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To calculate the entropy change when converting 20g of water at 30°C to steam at 250°C, three stages must be considered: heating the water to 100°C, vaporizing the water at 100°C, and heating the steam to 250°C. The heat capacity of liquid water is constant at 4.2 J/gK, while the heat of vaporization is 2260 J/g. The molar heat capacity of water vapor varies with temperature, defined by the equation provided. The total entropy change is obtained by integrating the heat capacity over the temperature range for each stage and incorporating the heat of vaporization. This comprehensive approach ensures an accurate calculation of the overall entropy change for the process.
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Homework Statement



At constant atmospheric pressure, 20g of water at 30C is converted into steam at 250C. Assume the heat capacity of liquid water is constant at 4.2 J/gK and the heat of vaporization at 100C is 2260 J/g. The molar heat capacity of water vapor at constant pressure is given by

\frac{c_p}{R} = a + bT + CT^2

where a= 3.634,{ } b= 1.195*10^{-3} K^{-1},{ } c=1.350*10^{-7} K^{-2}

Find the entropy change of the water.

Homework Equations



\Delta S = \frac{dQ}{T}
dQ = \int c_p dT

The Attempt at a Solution


My first thought was just to plug in the given equation for C_p into (1) and integrating, but I'm not sure how the heat of vaporization comes into play.
 
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There are 3 different 'stages' to this problem. You need to think of what they are, then find the entropy change that happens in each stage.
 

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