= Solution
{c}
At the <lifting condensation level>, the parcel <temperature> first equals its <dew point>. For $z/H\ll1$, the first-order <Taylor expansion> gives $T_D\simeq T_{D,s}(1-z/H)$. Equating this with the dry-<adiabatic process> result gives
$$
T_s-\Gamma_d z=T_{D,s}-\frac{T_{D,s}}H z,\qquad
\boxed{z_{\rm LCL}\simeq\frac{T_s-T_{D,s}}{\Gamma_d-T_{D,s}/H}.}
$$
Near terrestrial surface <temperatures>, $T_{D,s}/H\simeq1.7$–$1.9\,\mathrm{K\,km^{-1}}$, whereas $\Gamma_d\simeq9.8\,\mathrm{K\,km^{-1}}$. Consequently
$$
\boxed{z_{\rm LCL}\simeq125\,\mathrm{m\,K^{-1}}(T_s-T_{D,s}).}
$$
Use absolute <temperature> for $T_{D,s}/H$. The <temperature> difference can be evaluated in <kelvins> or degrees Celsius, whose increments agree. The coefficient is an approximate near-surface terrestrial value, not independent of composition or ambient conditions for all <atmospheres>.
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