It is not direct sunlight that is melting your ice mate. Let’s say the scoop has 10 cm² getting blasted from the sun, that’s 1 Watt of heat under maximum possible conditions (Sun vertically above you, perfectly black ice, etc.). tl;dr: In total from convention 1.8 W and condensation 2.5 W and radiation 0.65 W = 4.95 W -> maximum possible sunlight on earth would only increase this by 20 %, more realistic sunlight something like 10 %.

Actual math: Compare that to ambient temperatures of say, 30 °C, and let’s again say 10 cm² cross section, which translates to a diameter of 3.57 cm, so a sphere with a surface of 40 cm². The heat transfer coefficient under normal conditions is about 15 W/(m²K), so we get: 15 W/(m²K) * 0.004 m² * 30 K = 1.8 W

Additionally, we have latent heat from water (humidity) condensing on the cold surface: Let’s assume a Schmidt number of 0.6, so we get a mass transfer coefficient of: 15 W/(m²K) / [1.2 kg/m³ * 1000 J/(kgK)] * 0.6^(-2/3) = 0.0176 m/s Specific gas constant: 8.314 J/(molK) / 0.018 kg/mol = 462 J/(kgK) So the mass flux (condensation speed) is: 0.0176 m/s * 2000 Pa / [462 J/(kgK) * 273 K] = 0.00038 kg/(m²s)

Given the heat of condensation of 2257 kJ/kg water we thus get: 0.00038 kg/(m²*s) * 2257000 J/kg = 632 W/m²

And thus for our little sphere: 632 W/m² * 0.004 m² = 2.5 W

… Then we also have radiation from the hot surrounding, let’s assume 30 °C again, we get: Q = 5.67E-8 W/(m²*K^4) * 0.004 m² * (303 K^4 - 273 K^4) = 0.65 W (omitting radiation from the sky)