Chemical Thermodynamics: Entropy Change from Heat Transfer (Isothermal)

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Frances Docherty

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Created this as a copy of Entropy Change from Heat Transfer (Isothermal).

Name	Status	Author	Last Modified
Chemical Thermodynamics: Enthalpy Change and Vaporisation	Ready to use	Frances Docherty	30/03/2025 17:56
Chemical Thermodynamics: Entropy Change from Heat Transfer (Isothermal)	Ready to use	Frances Docherty	30/03/2025 17:55

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			Name	Type	Generated Value

entro	number	162.7651217596
trueheat	number	51800
ktemp	number	318.25
heat	number	51.8
ctemp	number	45.1

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162.7651217596

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Calculate the change in entropy when $\mathrm{\var{heat}\space kJ}$ of energy is transferred to a piece of zinc at $\mathrm{\var{ctemp}°C}$ (assume that the mass of zinc is sufficiently large such that its temperature can be considered constant):

Unnamed gap $\mathrm{J\space K^{-1}}$ .

Advice

Mathematically, the definition of entropy is:

$\mathrm{dS=\frac{dQ}{T}}$ ,

where $\mathrm{dS}$ is an infinitesimal entropy change; $\mathrm{dQ}$ is an infinitesimal heat transfer to the system being described (i.e. it has negative value if heat is removed from the system); and $\mathrm{T}$ is the temperature of the system, in $\mathrm{K}$ .

This equation works because the temperature of the system effectively remains constant during an infinitesimal heat transfer, whereas it generally does not work for any macroscopic heat transfer, as the temperature will change due to the heat transfer (the equation would effectively change as we added heat) - however, if the system is large enough, to distribute the heat over a large enough number of particles, then its temperature will remain effectively constant for fairly small heat transfers. This allows us to integrate the formula whilst treating $\mathrm{T}$ as a constant:

$\mathrm{\Delta S=\int dS=\int \frac{dQ}{T}=\frac{\Delta Q}{T}}$ .

You don't need to remember that explanation; just that if a system is large enough to assume a constant temperature, then the entropy change corresponding to a given heat transfer to the system is given by:

$\mathrm{\Delta S=\frac{\Delta Q}{T}}$ .

Note: If you are asked a question of this type and are not told that the system is large enough to assume constant temperature, but you are only told the heat transfer and one temperature, and expected to come to an answer, then it's probably safe to make the assumption yourself that the temperature remains constant and the entropy change will be given by the above formula.

We want to calculate the entropy change of some zinc when $\mathrm{\var{heat}\space kJ}$ of heat is transferred to it at $\mathrm{\var{ctemp}°C}$ . Let's convert our heat transfer to $\mathrm{J}$ by multiplying by one thousand, and convert our temperature to $\mathrm{K}$ by adding $\mathrm{273.15}$ to it. This gives us a heat transfer of $\mathrm{\var{trueheat}\space J}$ at a temperature of $\mathrm{\var{ktemp}\space K}$ .

We can now substitute our values into our equation to calculate the entropy change:

$\mathrm{\Delta S=\frac{\Delta Q}{T}}$

$\mathrm{\Rightarrow\space\space\space \Delta S=\frac{\var{trueheat}}{\var{ktemp}}=\var{sigformat(entro,3)}\space J\space K^{-1}}$ .

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Chemical Thermodynamics: Entropy Change from Heat Transfer (Isothermal)

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Frances Docherty 4 weeks, 1 day ago

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Frances Docherty 4 weeks, 1 day ago

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Chemical Thermodynamics: Entropy Change from Heat Transfer (Isothermal)

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Frances Docherty 4 weeks, 1 day ago

Frances Docherty 4 weeks, 1 day ago

Frances Docherty 4 weeks, 1 day ago

Error in variable testing condition

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