# Calculating the change in entropy, did I calculate correctly?

1. Apr 2, 2014

### PhyIsOhSoHard

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
$ΔS=\frac{Q}{T}$
$Q=mcΔT$

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

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

Total change of entropy:
$ΔS=-284 J/K+349 J/K=+65 J/K$

2. Apr 2, 2014

### collinsmark

Hello PhyIsOhSoHard,

Your approach looks correct to me.

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.