= Solution
With the standard liquid-water approximation $\rho\simeq10^3\,\mathrm{kg\,m^{-3}}$, each portion has <mass> $100\,\mathrm{kg}$. The equal <heat capacities> give final <temperature> $T_f=310\,\mathrm K$ by <energy> conservation. For an insulated mixing process with no other changes,
$$
\boxed{\Delta S_{\rm universe}=mc\ln\frac{T_f}{280}
+mc\ln\frac{T_f}{340}
=4.20\times10^5\ln\frac{310^2}{280\times340}
\simeq3.95\times10^3\,\mathrm{J\,K^{-1}}.}
$$
The reversible integrals $mc\int dT/T$ evaluate <entropy> differences between <equilibrium> states; the actual <heat> transfer is irreversible. These are two portions of the same substance, so there is no additional ideal-mixture compositional <entropy>.
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