I'm struggling with the following problem:

(Stephen J. Blundell, Concepts in Thermal Physics S. 154 Problem 14.5):

A block of lead of heat capacity 1kJ/K is cooled from 200K to 100K in two ways:

a) It is plunged into a large liquid bath at 100K.

b) It is first cooled to 150K in one liquid bath and then to 100K in another.

Calculate the entropy changes in the system comprising block plus baths in cooling from 200K to 100K in these two cases. Prove that in the limit of an infinite number of intermediate baths the total entropy change is zero.

This is my solution so far:

Bath: $dS = \frac{dQ}{T_{bath}} \Rightarrow \Delta S = \frac{\Delta Q}{T_{bath}} = \frac{C* \Delta T}{T_{bath}}$

Block: $dS = \frac{dQ}{T}; C = \frac{dQ}{dT}$

$\Delta S = \int_{T1}^{T2}\frac{C}{T}* dT = C * ln(\frac{T2}{T1})= C* ln(\frac{T2 * T_{m}}{T_{m}*T1}) = C*(ln(\frac{T2}{T_{m}})+ ln(\frac{T_{m}}{T1}))$

in both cases:

$\Delta S_{Block} = 693,1 \frac{J}{K}$

a) $ \Delta S_{bath} = C * \frac{100K}{100K} = 1000 \frac{J}{K}$

b) $\Delta S_{bath} = C* (\frac{50K}{150K} + \frac{50K}{100K} ) = \frac{5}{6}*1000 \frac{J}{K}$

Is this so far correct? I'm struggling with the infinite case:

$ \frac{\Delta S}{C} = \sum_{i = 1}^N \frac{100K}{n*(200K- i * \frac{100K}{n})}= \sum_{i= 1}^N \frac{1}{2n-i}$

I could not find any solution for $N \rightarrow \infty$. Where did I go wrong?

  • $\begingroup$ Shouldn't the change in entropy for the block be negative? $\endgroup$ – lionelbrits Dec 4 '13 at 0:28
  • $\begingroup$ Yes I might have messed up the signs at some point. $\endgroup$ – Stein Dec 4 '13 at 13:32

For the limit of an infinite number of intermediate baths, we can make the approximation that $T_\mathrm{bath} = T_\mathrm{block}$, so for the bath:

$\delta S_\mathrm{bath} = \frac{\delta Q}{T_\mathrm{block}} = -\frac{C\delta T_\mathrm{block}}{T_\mathrm{block}}$


$d S_\mathrm{bath} = -\frac{C}{T_\mathrm{block}} d T_\mathrm{block}$

Note that

$\sum_{i= 1}^N \frac{1}{2N-i} = \sum_{i= N}^{2N-1} \frac{1}{i} \approx \int_N^{2N-1} \frac{1}{x} dx \approx \ln 2$

  • $\begingroup$ where can i find more info on the series identity you've given/ $\endgroup$ – inya Nov 19 '16 at 18:41
  • $\begingroup$ The first identity is just relabling the indices on the sum. Then I recognize that it becomes an integral in the limit of infinite baths, and do the integration. $\endgroup$ – lionelbrits Dec 4 '16 at 18:48

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