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Consider a heat reservoir which gains heat $Q$ irreversibly at temperature $T$ from the surroundings which is at temperature $T_0$. The entropy change of reservoir is then given by $\frac{Q}{T}$, while that of the surroundings is $-\frac{Q}{T_0}$.
My question is, how is this possible? According to the Clausius inequality, the entropy change of a irreversible process is greater than that due to heat transfer. Please help, thank you!
The equation:
$$ dS = \frac{dQ}{T} $$
only applies to reversible processes. For an irreversible process $dS \gt dQ/T$.
To see this start with the expression for the change in internal energy:
$$ dU = dQ - dW $$
The internal energy is a state function, so this equation always applies whether the process is reversible or irreversible. So for a reversible process we have:
$$ dU = TdS - dW_{rev} $$
Suppose we make the same change in $U$ with an irreversible process then we have:
$$ dU = dQ_{irrev} - dW_{irrev} $$
And because $dU$ is the same in both cases we equate the two expressions to get:
$$ TdS - dW_{rev} = dQ_{irrev} - dW_{irrev} $$
which rearranges to:
$$ dS = \frac{dQ_{irrev}}{T} + \frac{dW_{rev} - dW_{irrev}}{T} $$
But we know that the work from a reversible process is always greater than the work from an irreversible process i.e. $dW_{rev} - dW_{irrev} > 0$, and this means:
$$ dS = \frac{dQ_{irrev}}{T} + \Delta $$
for some positive number $\Delta$ that depends on the details of the irreversible process.
The change of entropy is Q/T only if the heat transfer is reversible if the process is irreversible you can't obtain the change of entropy through the formula Q/T
After the heat transfer the change of entropy o the whole system is $\Delta S>Q/T-Q/T_0>0$
