= Solution
<Young's equation> gives $\gamma_{SV}-\gamma_{SL}=\gamma_{LV}\cos\theta$. The <interfacial-energy difference from two contact angles> follows by subtracting the two equilibrium contact-angle equations, which eliminates the common solid–air energy:
$$
\boxed{\gamma_{S,\mathrm{water}}-\gamma_{S,\mathrm{decane}}
=\gamma_{\mathrm{decane},V}\cos25^\circ-\gamma_{\mathrm{water},V}\cos102^\circ
\simeq3.67\times10^{-2}\,\mathrm{J\,m^{-2}}.}
$$
The water–solid interface has the larger energy. This assumes the same clean solid surface and equilibrium angles, without swelling or strong <contact-angle hysteresis>.
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