Change in Entropy of Boiling Water to Steam

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To find the change in entropy when boiling 2.00 kg of water at 100 degrees C to steam at the same temperature, the formula deltaS=Q/T is used, where Q is calculated as Q=m*Lvaporization. For this scenario, Q equals 4.52x10^6 J, derived from the mass and the latent heat of vaporization. The temperature T must be converted to an absolute scale, which is 373.15 K for 100 degrees C. The discussion clarifies that the calculation is straightforward once the correct temperature scale is applied. Understanding the use of absolute temperature is crucial for accurately determining entropy change.
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Homework Statement



Find the change in entropy when 2.00 kg of water at 100 degrees C is boiled away to steam at 100 degrees C. Give your answer in kJ/K

Homework Equations


deltaS=Q/T
Q=m*Lvaporization


The Attempt at a Solution


Q=m*Lvaporization
Q=2 kg * 22.6*10^5 J/kg
Q = 4.52x10^6 J

I'm not not sure how to find T though since it's at 100 degrees and going to 100 degrees...

Thanks!
 
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There's a good deal of views on here, does no one know how to do this? ;)
 
T is given. All you have to do is put it in the correct scale. The denominator in \Delta Q/T is the absolute temperature at which the heat flows into the water.

AM
 
Oh, I thought I had to do something more complicated with the T. Thanks!
 
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