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I did a few heat transfer calculations to help answer this question.

I will calculate convective heat loss from a cylinder approximately at 30cm wide, 2m tall with a temperature of 36C in the Martian atmosphere.   Wind speed = 10mph (4.5m/s).  Density of Martian CO2 at -50°C and 0.81KPa is 0.44mol/m3 or 0.02kg/m3.  Dynamic Viscosity is ~1x10-5pa.s.  On this basis, the Reynolds number is:

Re=ρvL/μ

L is characteristic length, which is equal to 4 times cross-sectional area, divided by perimeter or 0.15m in this case.  Reynolds is therefore 0.02x4.5x0.15/0.00001 = 1350.

Using the cylindrical cross-flow equation, Nusselt number can be calculated:

Nu = 0.193Re^0.618 x Pr^0.33

Prandtl, Pr=Cp.μ/k.

For CO2 under these conditions, k is 0.01W/m.K and Cp at 220K is 1960J/KgK.  So Prandtl is 1.96.  On this basis, Nusselt can be calculated to be 20.8.
The heat transfer coefficient, h= Nu.k/L.  Which works out to be 1.4W/m2K.

For a cyclindrical body with a surface area of ~1m2 and a temperature difference of 86K with the environment, heat loss rate can be calculated using Newton’s Law of Cooling:

Q=hA(T1-T2) = 1.4 x 1 x 86 = 120watts.

By contrast, radiation would lose 400watts of heat from the same body under the same conditions.  So on the Martian surface, heat loss will be dominated by radiation and conduction.  As a rough rule of thumb ‘the convection heat losses in the Martian atmosphere for any structure at room temperature, will be about 30% of the black body radiation heat loss rate’.

In a dust storm, with air velocity of 20m/s, the convective heat transfer coefficient increases roughly 2.5 times and thus reaches 75% of the black body radiation heat loss.