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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
    [itex]ΔS=\frac{Q}{T}[/itex]
    [itex]Q=mcΔT[/itex]

    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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