= Solution
The units supplied for 130 indicate <specific heat capacity>, so the block's total <heat capacity> is $C=mc=0.5(130)=65\,\mathrm{J\,K^{-1}}$. Treat it as constant, and the lake as a <thermal reservoir> at $T_0=283.15\,\mathrm K$. The initial block <temperature> is $T_i=373.15\,\mathrm K$.
Compute the block's <entropy> change using a reversible comparison path, even though the actual cooling is irreversible:
$$
\Delta S_{\mathrm{block}}=\int_{T_i}^{T_0}\frac{C\,dT}{T}
=C\ln\frac{T_0}{T_i}=-17.940\,\mathrm{J\,K^{-1}}.
$$
The lake receives $Q=C(T_i-T_0)=5850\,\mathrm J$ at constant <temperature>, so
$$
\Delta S_{\mathrm{lake}}=\frac Q{T_0}=20.660\,\mathrm{J\,K^{-1}}.
$$
Therefore
$$
\boxed{\Delta S_{\mathrm{Universe}}=2.72\,\mathrm{J\,K^{-1}}.}
$$
This is <entropy generated by cooling a finite body in a reservoir>. The positive total is consistent with the <Second law of thermodynamics>, although the cooling block's own <entropy> decreases.
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