= Solution
Take $x$ eastward, $y$ northward and $z$ upward, and let $u$ be the eastward <geostrophic wind>. With <Coriolis parameter> $f$ independent of $z$, <geostrophic balance>, <hydrostatic balance>, and the <ideal gas law> give
$$
fu=-\frac{p_y}{\rho},\qquad p_z=-\rho g,\qquad p=\rho R_dT.
$$
Here $R_d$ is the specific gas constant for dry air. Taking a vertical <derivative> of the first equation and interchanging mixed <partial derivatives> gives
$$
f u_z=g\frac{\rho_y}{\rho}-fu\frac{\rho_z}{\rho}.
$$
It matters that $\rho$ is not constant in the <pressure gradient force>. Substituting
$$
\frac{\rho_y}{\rho}=-\frac{fu}{R_dT}-\frac{T_y}{T},\qquad
\frac{\rho_z}{\rho}=-\frac g{R_dT}-\frac{T_z}{T}
$$
cancels the two terms proportional to $gfu/(R_dT)$ and yields the height-coordinate <thermal wind> relation
$$
f u_z=-\frac gT T_y+\frac{fu}{T}T_z.
$$
If $|uT_z|\ll|Tu_z|$, this reduces to
$$
\boxed{f\frac{\partial u}{\partial z}\simeq-\frac gT\frac{\partial T}{\partial y}.}
$$
The discarded term has relative scale $H|T_z|/T$ when $u_z\sim u/H$. Taking $H=10\ \mathrm{km}$, $T=250\ \mathrm K$ and a <atmospheric lapse rate> of $6.5\ \mathrm{K\,km^{-1}}$ gives about $0.26$. Thus this height-coordinate formula is a useful estimate, with potentially tens-of-percent compressibility-related correction through the vertical <temperature gradient>; it is not an exact consequence of the two balances. If the <temperature gradient> vanishes vertically, the displayed approximation is exact within those balances.
The <hydrostatic approximation> is very good on the broad, shallow atmospheric scales considered. The <geostrophic balance> is normally justified by a small <Rossby number> and fails as a general approximation near the equator. At $30^\circ\mathrm N$, $f=7.292\times10^{-5}\ \mathrm{s^{-1}}$. With the ideal $134\ \mathrm{m\,s^{-1}}$ wind and a $1000$--$2000\ \mathrm{km}$ horizontal scale, $\mathrm{Ro}=U/(fL)\simeq0.9$--$1.8$: quantitative accuracy is limited for this especially strong ideal jet. Small vertical aspect ratio alone does not ensure <geostrophic balance>, although a straight, steady zonal flow can satisfy it even when this scale estimate of the <Rossby number> is not small.
For the requested estimate, suppose the upper <jet stream> is $H=10\ \mathrm{km}$ above a weak surface wind, has <temperature> $T\simeq250\ \mathrm K$, and spans a meridional width $L$. Integrating the approximate <thermal wind> relation in scale form gives
$$
\boxed{T_{\rm south}-T_{\rm north}\simeq\frac{fTUL}{gH}
\simeq25\ \mathrm K\left(\frac{L}{1000\ \mathrm{km}}\right)
\left(\frac{10\ \mathrm{km}}H\right)
\left(\frac{T}{250\ \mathrm K}\right).}
$$
The northward <temperature gradient> is negative because the eastward wind strengthens upward. A $1000$--$2000\ \mathrm{km}$ jet width gives roughly $25$--$50\ \mathrm K$. No unique numerical difference follows without specifying the jet width, height and representative <temperature>; these are explicitly estimated quantities.
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