# Change in system entropy in relation to heat transfer

• tracker890 Source h
tracker890 Source h
Homework Statement
Establishing the Relationship Between Input Heat for Energy Conservation and Input Heat Defined by Entropy.
Relevant Equations
entropy balance equation.
entropy definition formula.

Q： What are the differences between the heat transfer calculated by the energy conservation equation and the heat transfer determined by the Gibbs relation? ##why\ \left( ans\_1 \right) \ne \left( ans\_2 \right) ##
reference

Energy balance：
$$Q_{in,net}-W_{out,net}=\cancel{\bigtriangleup U_{cv}}\cdots \left( 1 \right)$$
$$Q_{in,net}=-Q$$
$$W_{out,net}=-W_{in}$$
$$\therefore \left( 1 \right) =-Q+W_{in}=0$$
$$\therefore Q=W_{in}=200KJ\cdots \cdots \left( ans\_1 \right)$$

entropy balance equation.
entropy definition formula.
Gibbs relation.
$$dS_{sys}=\left( \frac{\delta Q}{T_k} \right) _{int.rev}\cdots \left( 2 \right)$$
$$\bigtriangleup S_{sys}=m\left( \frac{\cancel{du}+p\cancel{dv}}{T_k} \right) =0$$
$$\therefore \left( 2 \right) =0$$
$$\therefore Q=0KJ\cdots \cdots \left( ans\_2 \right)$$

Last edited:
tracker890 Source h said:
Homework Statement: Establishing the Relationship Between Input Heat for Energy Conservation and Input Heat Defined by Entropy.
Relevant Equations: entropy balance equation.
entropy definition formula.

View attachment 330839
Q： What are the differences between the heat transfer calculated by the energy conservation equation and the heat transfer determined by the Gibbs relation? ##why\ \left( ans\_1 \right) \ne \left( ans\_2 \right) ##
reference

Energy balance：
$$Q_{in,net}-W_{out,net}=\cancel{\bigtriangleup U_{cv}}\cdots \left( 1 \right)$$
$$Q_{in,net}=-Q$$
$$W_{out,net}=-W_{in}$$
$$\therefore \left( 1 \right) =-Q+W_{in}=0$$
$$\therefore Q=W_{in}=200KJ\cdots \cdots \left( ans\_1 \right)$$

entropy balance equation.
entropy definition formula.
Gibbs relation.
$$dS_{sys}=\left( \frac{\delta Q}{T_k} \right) _{int.rev}\cdots \left( 2 \right)$$
$$\bigtriangleup S_{sys}=m\left( \frac{\cancel{du}+p\cancel{dv}}{T_k} \right) =0$$
$$\therefore \left( 2 \right) =0$$
$$\therefore Q=0KJ\cdots \cdots \left( ans\_2 \right)$$
The entropy change of the system is zero since the gas temperature and volume are constant. The entropy transferred from the system to the surroundings is ##200/(30 + 273)=0.660\ kJ/K##. From the Clausius relationship, $$\Delta S=\frac{Q}{T_{surr}}+\sigma$$where ##\sigma## is the amount of entropy generated within the system during the process. So, $$0=-0.660+\sigma$$and the amount of generated entropy is equal to 0.660 kJ/K.

This question is a little ambiguous since, if the system is at 40 C and the surroundings are at 30 C, the system should be transferring heat to the surroundings until it too is at 30 C. But its final temperature is stated to be 40 C. It isn't clear what one should take as the temperature at the interface between the system and surroundings when the heat transfer Q and entropy transfer ##Q/T_{interface}##is occurring.

Last edited:
tracker890 Source h
Chestermiller said:
The entropy change of the system is zero since the gas temperature and volume are constant. The entropy transferred from the system to the surroundings is ##200/(30 + 273)=0.660\ kJ/K##. From the Clausius relationship, $$\Delta S=\frac{Q}{T_{surr}}+\sigma$$where ##\sigma## is the amount of entropy generated within the system during the process. So, $$0=-0.660+\sigma$$and the amount of generated entropy is equal to 0.660 kJ/K.

This question is a little ambiguous since, if the system is at 40 C and the surroundings are at 30 C, the system should be transferring heat to the surroundings until it too is at 30 C. But its final temperature is stated to be 40 C. It isn't clear what one should take as the temperature at the interface between the system and surroundings when the heat transfer Q and entropy transfer ##Q/T_{interface}##is occurring.
You've provided a detailed explanation, thank you. I finally understand.

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