# How much hotter will an object get if I paint it black?

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.

• Are you asking how to determine the emissivity of an object, or are you asking how to do the calculation assuming that you already have the emissivity as a function of wavelength? Commented Jul 13, 2020 at 0:01
• If what you have is emmisivity and absorbance at a given frequency, then you need to find the power emitted as an integral( Wikipedia) and set that equal to the absorbed power from the sun( if you want, integrated over absorbance as a function of frequency). Commented Jul 13, 2020 at 5:57
• Actually rereading the title, thermal conduction with the painted object may be much more important for you than radiative loss. Commented Jul 13, 2020 at 5:58
• this measurement may help phys.org/news/2011-10-silver-white-cars-cooler.html Commented Jul 13, 2020 at 6:03
• @probably_someone I am asking what the temp diff will be if I have two identical objects, one white and one black in the same light. Wikipedia lists e for snow as .8-.9 but for asphalt .88. so the Stefan-Boltzmann law shows the same result yet we all know that black objects get warmer Commented Jul 14, 2020 at 13:28

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 $$h$$).

• You actually came closest to answering my question so i gave you the check mark. However, the actual values of emissivity used in tables like the one above as well as this. nuclear-power.net/nuclear-engineering/heat-transfer/… Lead to temp differences smaller than i found. I must be doing something wrong. Commented Jul 15, 2020 at 0:12
• I tested yesterday porcelain tiled floor black and white. The temperature differences were about 10 K after some hours under the sun. As you can see from the equation, that depends also on $h$, that is another empirical factor. Commented Jul 15, 2020 at 0:27
• Convection...hmmm.. I have to read how emissivity is measured. They would have to do it in a vacuum to eliminate this effect. The convective coefficient can be 2-20 for air. nuclear-power.net/nuclear-engineering/heat-transfer/… Commented Jul 15, 2020 at 0:48
• @ClaudioSaspinski I am not sure how your equation works. The LHS of your equation has units of kelvin while the RHS has units of kelvin multiplied by the units of $h$. How can we have different units on either side of an equation? From my understanding, the governing equation should be something like $P_{in,rad}=P_{out,rad}+P_{out,convec}$. Am I correct? Commented Aug 18, 2022 at 14:28
• @SalahTheGoat You are right, it doesn't make sense. The idea must be to equate the incoming radiating energy with the heat loss, that due to the low object temperature, is mainly convective. I will try to work on it. Commented Aug 18, 2022 at 15:49

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.

• yes your thinking is what I went through. But I wanted to CALCULATE the actual temperature. I did the experiment (a piece of paper half sprayed black in a big cardboard box whose top was open) and used an infrared thermometer which read the black paper as hotter (just as your reference found with cars). The fact that emissivity and absorbance are the same and the conclusion that they should have the same temperature at equilibrium is not agreeing with experience so I assume that more is needed for the Stefan-Boltzmann law to agree with reality. efharisto. Commented Jul 14, 2020 at 18:44
• well, your experiment is measuring the heat by the radiation, not by the standard thermometer, If the standard thermometer could be used, as in the figure? If you are serious in experimenting I would use two boxes of different colors and a standard thermometer inside. Commented Jul 15, 2020 at 3:48
• I mean that in real life the black body curve is not exactly followed. It could be that the emissivity of the black or the white side is affected differently (the curve is different) Commented Jul 15, 2020 at 4:12
• Sometimes, especially near ambient temperatures, readings may be subject to error due to the reflection of radiation from a hotter body—even the person holding the instrument[citation needed] — rather than radiated by the object being measured, and to an incorrectly assumed emissivity. en.wikipedia.org/wiki/Infrared_thermometer Commented Jul 15, 2020 at 4:22