Is Dropping a Metal Block into a Lake a Reversible Process in Thermodynamics?

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Sum Guy
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Hello all.

I have a quick question about entropy... I've just been formally introduced to it.

Consider the example of a metal block of mass m and heat capacity Cp at temperature T1 = 60C being dropped into a large lake of temperature T2 = 10C.

$$\Delta S_{block} = \int\frac{dQ_{reversible}}{T} = \int_{T_1}^{T_2}\frac{C_pdT}{T}$$

I have a few questions... how would the block reaching thermal equilibrium with the lake be classed as a reversible process (it must be otherwise we wouldn't have the above calculation)? Why can we ignore the blocks mass? Are there any general tips you could provide that might help me tackle questions on entropy?

Many thanks.
 
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Sum Guy said:
Hello all.

I have a quick question about entropy... I've just been formally introduced to it.

Consider the example of a metal block of mass m and heat capacity Cp at temperature T1 = 60C being dropped into a large lake of temperature T2 = 10C.

$$\Delta S_{block} = \int\frac{dQ_{reversible}}{T} = \int_{T_1}^{T_2}\frac{C_pdT}{T}$$

I have a few questions... how would the block reaching thermal equilibrium with the lake be classed as a reversible process (it must be otherwise we wouldn't have the above calculation)? Why can we ignore the blocks mass? Are there any general tips you could provide that might help me tackle questions on entropy?

Many thanks.
You are correct in recognizing that the process you described is not a reversible process. But, to calculate the change in entropy of the block for this change, you need to devise a reversible path from the initial state to the final state, and calculate the integral of dQ/T for that path. One such path is where you contact the block with a sequence of reservoirs having gradually lower temperatures (so that its temperature only differs only slightly from that of the reservoir it is currently in contact with). That leads to the integral you wrote down. Of course, that integral only gives the entropy change per unit mass. To get the actual entropy change of the block, you need to multiply by the mass of the block.
 
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