The Entropy Change of Melting Ice: Why is the Equation Written as ΔS = Q/T?

[ChiralSuperfields]

Homework Statement

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Relevant Equations

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For this,

Why dose they write the change in entropy equation as $\Delta S = \frac{Q}{T}$? Would it not better to write it as $\Delta S = \frac{\Delta Q}{T}$, since it clear that we are only concerned about the transfer of heat in our system while it remains at constant temperature as all the energy goes into the change in phase.

Many thanks!

 

[TSny]

Homework Statement:: Please see below
Relevant Equations:: Please see below

For this,
View attachment 324174

Why dose they write the change in entropy equation as $\Delta S = \frac{Q}{T}$? Would it not better to write it as $\Delta S = \frac{\Delta Q}{T}$, since it clear that we are only concerned about the transfer of heat in our system while it remains at constant temperature as all the energy goes into the change in phase.

Many thanks!

You are thinking correctly. However, $Q$ is often used for the heat transferred in a process. The symbol $\Delta Q$ for the heat transferred is also sometimes used. But, this symbol can be misleading if $\Delta Q$ is interpreted as "the change in $Q$", or as $\Delta Q = Q_f - Q_i$. The symbols $Q_f$ and $Q_i$ do not have any meaning. The initial and final states of the system do not "have heat". In more formal words, "heat is not a state variable".

 

[ChiralSuperfields]

You are thinking correctly. However, $Q$ is often used for the heat transferred in a process. The symbol $\Delta Q$ for the heat transferred is also sometimes used. But, this symbol can be misleading if $\Delta Q$ is interpreted as "the change in $Q$", or as $\Delta Q = Q_f - Q_i$. The symbols $Q_f$ and $Q_i$ do not have any meaning. The initial and final states of the system do not "have heat". In more formal words, "heat is not a state variable".

Thank you for your reply @TSny! Oh I think I see what you mean. Since heat is a 'transfer variable', it inherently describing a transition between the of energy between the finial and previous state.

 

[TSny]

Thank you for your reply @TSny! Oh I think I see what you mean. Since heat is a 'transfer variable', it inherently describing a transition between the of energy between the finial and previous state.

Yes. $Q$ is always associated with a process, and it refers to the energy transferred due to a temperature difference during the process. Likewise, work $W$ is always associated with a process. You might sometimes see $\Delta W$ for the work, but we would never think of this as $\Delta W = W_f - W_i$.

 

[ChiralSuperfields]

Yes. $Q$ is always associated with a process, and it refers to the energy transferred due to a temperature difference during the process. Likewise, work $W$ is always associated with a process. You might sometimes see $\Delta W$ for the work, but we would never think of this as $\Delta W = W_f - W_i$.

Ahh thank you very much @TSny! That is a very good analogy!