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 convenction 1.8 W, 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)
Not quite sure how this would affect melting ice cream. It does fill in some missing pieces to climate models. There are more clouds around than the models predict, which raises the planet’s albedo.
Yes, while the radiation puts more energy in than the convective etc. cooling removes. So near 0 this is guaranteed, since the temperature difference from ice to ambient is almost 0 while radiation keeps pumping in something like 0.5 W. But who eats ice at freezing temperatures… And outside?
I have eaten ice cream outside when temperatures were sub-zero Fahrenheit. It’s not something I do regularly but it’s happened and will probably happen again.
If I want ice cream, then I want ice cream. No other considerations matter.
I was just sharing my anecdote as a counterpoint to your minor rhetorical point at the end, because at least to me, it’s funny since eating ice cream outside at -10 degrees is a ridiculous thing to do.
Though, I will note that while ice cream won’t melt at those temperatures, at atmospheric pressure it will still sublimate. So, in that way you could still lose your ice cream without intervention, it would just take a while.
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 convenction 1.8 W, 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)
The science on this is relatively recent, but it turns out there is a photomolecular effect on evaporating water that can’t be explained with heat.
https://news.mit.edu/2024/how-light-can-vaporize-water-without-heat-0423#%3A~%3Atext=Light%2C+striking+the+water's+surface%2Cwide+range+of+significant+implications.
Not quite sure how this would affect melting ice cream. It does fill in some missing pieces to climate models. There are more clouds around than the models predict, which raises the planet’s albedo.
So made this meme is eating ice cream when it’s below or near freezing? Because you still get ice melting below freezing due to radiation.
Yes, while the radiation puts more energy in than the convective etc. cooling removes. So near 0 this is guaranteed, since the temperature difference from ice to ambient is almost 0 while radiation keeps pumping in something like 0.5 W. But who eats ice at freezing temperatures… And outside?
I have eaten ice cream outside when temperatures were sub-zero Fahrenheit. It’s not something I do regularly but it’s happened and will probably happen again.
If I want ice cream, then I want ice cream. No other considerations matter.
Alright then. But at 0 °F it is not going to melt without your intervention, no matter how sunny it is.
Absolutely, I don’t disagree with that.
I was just sharing my anecdote as a counterpoint to your minor rhetorical point at the end, because at least to me, it’s funny since eating ice cream outside at -10 degrees is a ridiculous thing to do.
Though, I will note that while ice cream won’t melt at those temperatures, at atmospheric pressure it will still sublimate. So, in that way you could still lose your ice cream without intervention, it would just take a while.