= Solution
For <entropy gain from impact and thermal equilibration>, let the large lake be a reservoir at an initially unspecified <temperature> $T_L$, in kelvin, and take the final meteorite <temperature> to be $T_L$. Its initial <temperature> is $T_i=1273.15\,\mathrm K$. The meteorite loses thermal energy $mc(T_i-T_L)$, and its <kinetic energy> $mv^2/2$ is also dissipated into the lake. Thus
$$
\Delta S_{\rm met}=mc\log(T_L/T_i),\qquad\Delta S_{\rm lake}=\frac{mc(T_i-T_L)+mv^2/2}{T_L},
$$
and
$$
\boxed{\Delta S_{\rm universe}=mc\left[\log\frac{T_L}{T_i}+\frac{T_i}{T_L}-1\right]+\frac{mv^2}{2T_L}.}
$$
Both <heat> equilibration and dissipation contribute positively. For the requested estimate, assume a lake <temperature> $T_L=300\,\mathrm K$. This gives \b[$\Delta S_{\rm universe}\simeq2.93\times10^3\,\mathrm{J\,K^{-1}}$]. Taking $273$–$300\,\mathrm K$ instead gives about $3.3$–$2.9\,\mathrm{kJ\,K^{-1}}$; this shows the estimate's sensitivity without asserting an unprovided lake <temperature>. The boxed expression retains the full <temperature> dependence. The model uses the stated constant <heat capacity>, no phase change of the meteorite, and a lake large enough that its <temperature> change is negligible.<Classical Thermodynamics by David Tong>[Section 4.3.3: entropy changes and thermal reservoirs]{ref}
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