How much hotter will an object get if I paint it black? I have read these threads
If black is the best absorber and radiator, why does it get hot?
Black and white matters. But why and how?
If a black body is a perfect absorber, why does it emit anything?
Why is black the best emitter?
Some respondents referred to the Stefan-Boltzmann Law and indeed were kind enough to do the calculation. This post
Emissivity and Final Temperature of a Black and White object
indicates that the emissivity constant should be different for white objects than for black objects. Wikipedia shows for example
https://en.wikipedia.org/wiki/Emissivity
states that 'white paint absorbs very little visible light. However, at an infrared wavelength of 10x10−6 metres, paint absorbs light very well, and has a high emissivity. '
I am still at a loss though as to how to apply the Stefan-Boltzmann equation to calculate the equilibrium temperature of two identical objects (for example a piece of paper) in the identical sunlight(light intensity of 1000 W/m2 (typical for cloudless sunny day)) that differ only in color.
 A: When the objects are exposed to sunlight, they are heated by radiation, and cooled mainly by convection: $\frac{q}{A} = h(T_{obj} - T_{air})$, where $h$ is the convective coefficient.
In order to estimate the transfer of heat by radiation to an object on earth by the sunlight, we can imagine a cone with apex in the center of the Sun. The energy from the surface of that cone at the surface of the Sun is transferred to a given area of the object. On the other hand the object radiates according to its temperature and emissivity. Using the law of Stefan-Boltzmann, the net influx is:
$\frac{q}{A} = \sigma(fT_s^4 - \epsilon T_{obj}^4)$, where $f$ is the ratio between the area at the surface of the Sun to the corresponding area of the object, and $\epsilon$ is the emissivity of the object. Using the known values for the radius of the Sun and the distance Earth-Sun, $f = 2,15*10^{-5}$.
Testing this model to calculate only the solar input: $\frac{q}{A} = \sigma fT_s^4 = 5,67*10^{-8}*2,15*10^{-5}*5273^4 = 942\, Wm^{-2}$, that is close to the OP figure.
The thermal equilibrium is reached when:
$h(T_{obj} - T_{air}) = \sigma(fT_s^4 - \epsilon T_{obj}^4)$
Putting some numbers:
for $h = 10Wm^{-2}K^{-1}$, $T_{air} = 298 K$, $\epsilon = 1$ $\implies T_{obj} = 327 K$
for $h = 10Wm^{-2}K^{-1}$, $T_{air} = 298 K$, $\epsilon = 0,5$ $\implies T_{obj} = 350 K$
Of course, the magnitude of the difference is bigger for lower convective loss (smaller $h$).
A: In this link, a comparison is made between colors of cars, before reaching thermodynamic equillibrium.

Thermodynamic equilibrium is an axiomatic concept of thermodynamics. It is an internal state of a single thermodynamic system, or a relation between several thermodynamic systems connected by more or less permeable or impermeable walls. In thermodynamic equilibrium there are no net macroscopic flows of matter or of energy, either within a system or between systems.

If your two objects stay in the sunlight long enough to reach thermodynamic equillibrium, the zeroth law should say that their final temperatures are same:

The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third one, then they are in thermal equilibrium with each other.

See the explanation of thermal equilibrium here.


Figure 1.2.1: If thermometer A is in thermal equilibrium with object B, and B is in thermal equilibrium with C, then A is in thermal equilibrium with C. Therefore, the reading on A stays the same when A is moved over to make contact with C.

Emissivity and absorptivity would play a role to how long it would take for the two different colored objects to reach thermodynamic equilibrium with the air surrounding them at the same input  radiation.
The tests with cars show that the time is important in showing the differences in the color of the  car, and the particular case has to be taken into account. I would think that the two pieces of different color paper ( no wind)  should reach equilibrium in the noon sun  fairly soon, and thus the same temperature. In general one should use the emissivity and absroptivity to solve a specific case, but it is not simple calculations.
