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Homework Help: Calculating the change in entropy, did I calculate correctly?

  1. Apr 2, 2014 #1
    1. The problem statement, all variables and given/known data
    A 3.5 kg block of cobber at 100 degrees celsius (373 K) is put in 0.8 kg water at 0 degrees celsius (273 K).
    The equilibrium temperature is 30 degrees celsius (303 K).
    Calculate the change of entropy for the system of cobber and water.

    2. Relevant equations

    3. The attempt at a solution
    Change of entropy for the cobber block:
    [itex]ΔS=S_2-S_1=\int^2_1\frac{dQ}{T}=\int^{T_2}_{T_1}mc\frac{dT}{T}=mc \ln\frac{T_2}{T_1}=3.5kg\cdot 390\frac{J}{kg K}\left ( \ln\frac{303 K}{373 K} \right )=-284 J/K[/itex]

    Change of entropy for the water:
    [itex]ΔS=S_2-S_1=\int^2_1\frac{dQ}{T}=\int^{T_2}_{T_1}mc\frac{dT}{T}=mc \ln\frac{T_2}{T_1}=0.8kg\cdot 4190\frac{J}{kg K}\left ( \ln\frac{303 K}{273 K} \right )=349 J/K[/itex]

    Total change of entropy:
    [itex]ΔS=-284 J/K+349 J/K=+65 J/K[/itex]
  2. jcsd
  3. Apr 2, 2014 #2


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    Homework Helper
    Gold Member

    Hello PhyIsOhSoHard,

    Your approach looks correct to me. :approve:

    However, I would be careful with rounding numbers too early. For example, some of the sources I see online for the specific heat capacity for copper are around 386 J/(Kg K) rather than 390.

    As a rule of thumb, try to keep a couple of extra significant digits around in the intermediate calculations, then round to the appropriate number of significant figures at the very end. :smile:
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