Calculating the change in entropy, did I calculate correctly?

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The discussion focuses on calculating the change in entropy for a system consisting of a 3.5 kg block of copper and 0.8 kg of water reaching thermal equilibrium. The calculated change in entropy for the copper block is -284 J/K, while for the water, it is +349 J/K, resulting in a total change of +65 J/K for the system. One participant advises caution regarding rounding numbers too early, suggesting that specific heat values should be verified for accuracy. Maintaining extra significant digits during calculations before rounding at the end is recommended for precision. Overall, the calculations and methodology presented are deemed correct.
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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:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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