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This answer is somewhat hand-wavy, but I do believe it should help to grasp the concepts on an intuitive level.

First of all, entropy is not a measure of randomness. For an isolated system in equilibrium under the fundamental assumption of statistical mechanics, the entropy is just $$ S=k\ln\Omega $$ where $\Omega$ is the number of micro-statesmicrostates - microscopic system configurations - compatible with the given macro-statemacrostate - macroscopic equilibrium state characteristed by thermodynamical variables.

It follows from the second law $$ \delta Q = T\mathrm{d}S=T\mathrm{d}(k\ln\Omega)=kT\frac1\Omega\mathrm{d}\Omega $$ or equivalently $$ \mathrm{d}\Omega = \Omega\frac{\delta Q}{kT} $$ The energy $kT$ is related to the average energy per degree of freedom, so this formula tells us that the transfer of heat into a system at equilibrium opens up a new number of microstates proportional to the number of existing ones and the number of degrees of freedom the transferred energy may excite.

This answer is somewhat hand-wavy, but I do believe it should help to grasp the concepts on an intuitive level.

First of all, entropy is not a measure of randomness. For an isolated system in equilibrium under the fundamental assumption of statistical mechanics, the entropy is just $$ S=k\ln\Omega $$ where $\Omega$ is the number of micro-states - microscopic system configurations - compatible with the given macro-state - macroscopic equilibrium state characteristed by thermodynamical variables.

It follows from the second law $$ \delta Q = T\mathrm{d}S=T\mathrm{d}(k\ln\Omega)=kT\frac1\Omega\mathrm{d}\Omega $$ or equivalently $$ \mathrm{d}\Omega = \Omega\frac{\delta Q}{kT} $$ The energy $kT$ is related to the average energy per degree of freedom, so this formula tells us that the transfer of heat into a system at equilibrium opens up a new number of microstates proportional to the number of existing ones and the number of degrees of freedom the transferred energy may excite.

This answer is somewhat hand-wavy, but I do believe it should help to grasp the concepts on an intuitive level.

First of all, entropy is not a measure of randomness. For an isolated system in equilibrium under the fundamental assumption of statistical mechanics, the entropy is just $$ S=k\ln\Omega $$ where $\Omega$ is the number of microstates - microscopic system configurations - compatible with the given macrostate - macroscopic equilibrium state characteristed by thermodynamical variables.

It follows from the second law $$ \delta Q = T\mathrm{d}S=T\mathrm{d}(k\ln\Omega)=kT\frac1\Omega\mathrm{d}\Omega $$ or equivalently $$ \mathrm{d}\Omega = \Omega\frac{\delta Q}{kT} $$ The energy $kT$ is related to the average energy per degree of freedom, so this formula tells us that the transfer of heat into a system at equilibrium opens up a new number of microstates proportional to the number of existing ones and the number of degrees of freedom the transferred energy may excite.

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source | link

This answer is somewhat hand-wavy, but I do believe it should help to grasp the concepts on an intuitive level.

First of all, entropy is not a measure of randomness. For an isolated system in equilibrium under the fundamental assumption of statistical mechanics, the entropy is just $$ S=k\ln\Omega $$ where $\Omega$ is the number of micro-states - microscopic system configurations - compatible with the given macro-state - macroscopic equilibrium state characteristed by thermodynamical variables.

It follows from the second law $$ \delta Q = T\mathrm{d}S=T\mathrm{d}(k\ln\Omega)=kT\frac1\Omega\mathrm{d}\Omega $$ or equivalently $$ \mathrm{d}\Omega = \Omega\frac{\delta Q}{kT} $$ The energy $kT$ is related to the average energy per degree of freedom, so this formula tells us that the transfer of heat into a system at equilibrium opens up a new number of microstates proportional to the number of existing ones and the number of degrees of freedom the transferred energy may excite.