 Solution

Use the intended hot-ridge state and a linear law of thermal expansion about mantle density ${\rho }_m$ at $T_1$. Let $s$ be depth below the old seafloor. Its excess density isand its excess column mass is ${\int }_0^a\delta \rho ds={\rho }_m\alpha \Delta Ta/2$. Isostasy requires equal pressure at a compensation depth below both columns. Lowering the old seafloor by $h$ replaces rock by water, reducing column mass by $({\rho }_m-{\rho }_w)h$. Therefore the isostatic ridge elevation in a plate model isHere $\alpha \Delta T=0.052$; the result is a first-order estimate and neglects sediments, crustal-composition differences and flexural support. If one instead insists on the literal printed zero-age profile, its column is denser than the old plate and the formal relative elevation reverses sign, $-3.9\mathrm{km}$ to this order. That profile cannot describe the elevated hot ridge posited by the question.