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Change in Entropy

  1. Jan 19, 2005 #1
    >> Calculate the change in entropy of 210 g of water heated slowly from 25.0°C to 70.0°C. (Hint: Note that dQ = mcdT.) <<

    I'm not sure how to start this, but I think I need to use the equation:
    dS=m1*c1*ln(Tf/T1) + m2*c2*ln(Tf/T2)

    I know c of water = 4.186J and I know m1=m2... but I don't have a second temperature to use to solve the equation. But this is the only change in entropy equation I know of... am I missing something? :frown:
  2. jcsd
  3. Jan 19, 2005 #2
    Okay, I found part of my mistake but now I don't know why this isn't getting the correct answer...

    I used dQ = mcdT and plugged in:
    dQ = (.210kg)*(4.186J)*(45K)
    dQ = 39.5570J

    Then I used...
    dS = dQ/T
    Plugging in...
    dS = 39.5570/343.15K
    dS = .115J/K

    What am I doing wrong??
  4. Jan 19, 2005 #3


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    Homework Helper

    Remember that the formula dS=dQ/T is at a specific temperature for an infintesimal change in heat. What you did above is incorrect as it assumes that temperature is constant.



    so dS=mcdT/T

    Integrate both sides... what do you get?
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