Calculating the change in entropy, did I calculate correctly?

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SUMMARY

The change in entropy for a system consisting of a 3.5 kg block of copper at 100 degrees Celsius (373 K) and 0.8 kg of water at 0 degrees Celsius (273 K) reaching an equilibrium temperature of 30 degrees Celsius (303 K) is calculated to be +65 J/K. The entropy change for the copper block is -284 J/K, while the water contributes +349 J/K. This calculation utilizes the equations ΔS = Q/T and Q = mcΔT, with specific heat capacities of 390 J/(kg K) for copper and 4190 J/(kg K) for water. It is recommended to maintain precision in intermediate calculations to avoid rounding errors.

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Homework Statement


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.

Homework Equations


ΔS=\frac{Q}{T}
Q=mcΔT

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
 
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Hello PhyIsOhSoHard,

PhyIsOhSoHard said:

Homework Statement


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.

Homework Equations


ΔS=\frac{Q}{T}
Q=mcΔT

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

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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