Find change in entropy for a system with a series of reservoirs

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SUMMARY

The discussion focuses on calculating the change in entropy for a system interacting with a series of reservoirs. The user successfully derived the change in entropy of the material, represented as $$\Delta S_1 = C \ln{\frac{T_i+(t+1)\Delta T}{T_i+t\Delta T}}$$. However, they seek clarification on the change in entropy of the reservoir, which is given as $$\Delta S_2 = -\frac{C\Delta T}{T_i + (t+1)\Delta T}$$ in the reference material. The user expresses confusion regarding the heat capacity of the reservoir and its implications for entropy calculations.

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  • Basic understanding of ideal constant-temperature reservoirs.
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mcas
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Homework Statement
A material is brought from temperature ##T_i## to temperature##T_f## by placing
it in contact with a series of ##N## reservoirs at temperatures ##T_i + \Delta T##, ##T_i + 2\Delta T##, ..., ##T_i + N \Delta T = T_f##. Assuming that the heat capacity of the material,
C, is temperature independent, calculate the entropy change of the total
system, material plus reservoirs. What is the entropy change in the limit
##N \rightarrow \infty## for fixed ##T_f - T_i##?
Relevant Equations
##dS = \frac{1}{T} Q_{reversibile}##
I've calculated the change in the entropy of material after it comes in contact with the reservoir:

$$\Delta S_1 = C \int_{T_i+t\Delta T}^{T_i+(t+1)\Delta T} \frac{dT}{T} = C \ln{\frac{T_i+(t+1)\Delta T}{T_i+t\Delta T}}$$

Now I would like to calculate the change in the entropy of the reservoir. The answer in the book is:

$$\Delta S_2 = -\frac{C\Delta T}{T_i + (t+1)\Delta T}$$
And I don't know where this answer comes from. How am I supposed to find the change in entropy if I don't know what is the heat capacity of the reservoir?
 
Physics news on Phys.org
For an ideal constant-temperature reservoir, the change in entropy is always $$\Delta S=\frac{Q}{T_R}$$ where ##T_R## is the reservoir temperature.
 
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Chestermiller said:
For an ideal constant-temperature reservoir, the change in entropy is always $$\Delta S=\frac{Q}{T_R}$$ where ##T_R## is the reservoir temperature.

Thank you! I didn't know that.
 

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