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The Law of Thermodynamics says that two bodies will eventually have equal temperatures. How is it possible that when you leave your car in the sun, it gets hotter in the car than it is outside? Why isn’t the car at the same temperature as the outside, as it should be according to the Law?

TRiG
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bodacydo
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    That law is valid when there are no heat sources. For a car left in the sun, that's not the case. – Javier Jul 20 '15 at 01:56
  • What do you mean? Do you mean asphalt is heating it up through tires? What if I leave car on grass - it still gets hotter than outside. – bodacydo Jul 20 '15 at 02:00
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    It will have the same temperature as the outside, eventually – David says Reinstate Monica Jul 20 '15 at 03:30
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    The laws of thermodynamics do ensure there is a maximum possible temperature the car can reach, but it is related to the temperature of the sun's corona rather than the air temperature. – Harry Johnston Jul 20 '15 at 04:19
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    Or, in layman terms, your car works as a greenhouse. Parts inside are heated by the sun but the heat exchange is made slower by the envelope of the car (body, windows, etc). Laminated windscreens that reflect portion of the solar thermal radiation rather than let it pass through are one of the readily visible explanations to this. – Pavel Jul 20 '15 at 13:38
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    Question rephrased: Why is the inside of my pot on a stove with fire hotter than the outside? Now the question seems very silly. – Derek 朕會功夫 Jul 21 '15 at 00:48

4 Answers4

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Law of Thermodynamics says that two bodies eventually will have equal temperatures.

That is not an absolute Law. There are conditions, and one of those conditions involves the energy input to the bodies. If this Law was absolute, then the Sun would be at the same temperature as the universe, about 2.7 K, because the universe is much larger than the Sun. But the Sun has an internal energy converter/source which raises its local temperature.

The interior of a closed car in the sunlight will be higher because of a greenhouse effect. The glass of the car is transparent to the visible light, so that energy is absorbed by the interior of the car (the seats, dashboard, and floor) increasing their temperature. Those items then emit infrared radiation and the glass is fairly opaque to that radiation and the energy stays in the car. So more energy comes in the glass than is escaping out of the glass.

Because the trunk/boot doesn't have a glass opening to let radiation in, it will generally stay quite a bit cooler than the passenger compartment. Whatever radiation the trunk lid gets is reflected and radiated back out fairly efficiently. That's not to say it doesn't get hot, but it doesn't get to the same as the passenger compartment.

Bill N
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    @bodacydo - in other words, eventually power in will equal power out. In this case, power in (visible light) enters (is absorbed) very efficiently. But power out (infrared centered at about 10 um) is not efficiently radiated. So the inside has to get quite hot before the radiated power equals the absorbed power. – WhatRoughBeast Jul 20 '15 at 02:35
  • @Bill N. There is a question I have had for quite some time now and this question rises it again. If I place a windshield sun shade inside the car, the fact that the winshield surface is metallic or black makes a huge difference, doesn't it? if it is black more infrared will be emitted thus those kind of shades will be useless metallic ones will emit little infrared thus keeping the car cooler than without them. Is that asumption right or the light that enters the car has as much infrared as the one is emitted by the seats, dashboard, etc...? – YoMismo Jul 20 '15 at 11:32
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    @YoMismo The light-colored (silver or white) shiny windshield shades will reflect the visible light back out the glass rather than absorbing it. The issue is not so much the amount or rate of IR that enters as it is the total power absorbed by the interior, raising its temp, resulting in more trapped energy. If you can reflect energy, that helps keep the interior cooler. – Bill N Jul 20 '15 at 12:56
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    @YoMismo, the light color of your sun shade defeats the greenhouse effect. Most of the power in solar radiation is in short wavelengths that pass through glass. Most of the power in the black body radiation from a hot dashboard or hot upholstery is in longer wavelengths that are blocked by the glass. A light-colored shade reflects much of the short-wave radiation back out through the glass before it can be absorbed inside the car. – Solomon Slow Jul 20 '15 at 16:44
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    Convection has a must larger effect than radiation, as can be demonstrated by rolling the windows down, or even just cracking them a bit. – Michael Jul 20 '15 at 17:42
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    As @Michael notes the effect that drives car heating is the same as that which drives actual greenhouses. Trapping air and preventing convection. This is different from the greenhouse effect, ie. car heating is not mainly that light at some wavelengths are transmitted better than other. – Taemyr Jul 21 '15 at 14:15
  • If the IR wavelengths were transmitted back out through the glass as efficiently as as the visible, then the interior wouldn't get so hot. The difference in wavelength IS the main reason. I agree, if you allow ventilation, you can keep the car at air temperature, so it's both/and, not either/or. But the wavelength transmission effect is extremely important in this phenomenon. – Bill N Jul 21 '15 at 14:23
  • So you're saying I should wait in the trunk… – Neil G Aug 18 '15 at 11:51
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    @NeilG No, you should start riding a motorcycle. – Bill N Aug 18 '15 at 12:33
  • @BillN I'm very late to the party. But I would just like to confirm the above comments: preventing convection is indeed the main reason for the interior of the car heating up. The difference in wavelength is not specific to cars. Open surfaces will absorb visible light and emit IR light as well, the reason they won't heat as much as the car interior is the presence of convection. – Milan Nov 04 '19 at 18:51
  • @Milan Why does the air get warm? NOT because it absorbs the visible light, or even the infrared, but because it's in contact with the warmer surfaces. The surfaces remain warm, and continue to get hot is because the IR isn't emitted through the glass effectively. Remove the air, and the inside will still get warm and remain warm. The air is warming simply because it comes into equilibrium with the solid surfaces which absorb the primarily visible radiation. – Bill N Nov 04 '19 at 20:40
  • @BillN No. The surfaces remain hot because they are in contact with the warmer surfaces and because heat is not transported by convection. Meteorology Today, 12th edition, page 45: "Studies have shown that the warm air inside a greenhouse is caused more by the air’s inability to circulate and mix with the cooler outside air than by the entrapment of infrared energy". – Milan Nov 05 '19 at 14:21
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While respecting the other good and thoughrough answers, I feel I can give you a simple explanation to your exact question.

As you mention in your question, two bodies eventually will have equal temperatures when in thermal contact. "Eventually" is the key. If one body's temperature is raised, the bodies will eventually find a new equilibrium temperature.

But this law doesn't prevent the rise in temperature of one body by an external source. It only says what happens from then on. As long as energy is constantly added, this temperature equality that you seek is never reached.

  • The 1st law of thermodynamics is the law of energy conservation which explains the temperature rise.

$$\Delta U=Q-W$$

The Sun adds heat $Q$ to the car through radiation. No work is done $W=0$. Thus, internal energy $U$ must rise (corresponding to a rise in temperature).

The final steady temperature that the car reaches at a sunny day will both depend on the heat exchange with the outside air and on the incoming (as well as outgoing) radiation. All this balances at some point, by the temperature rising until heat leaving the car every second equals heat into the car every second. The temperature will not be constant until this equilibrium state is reached.

Steeven
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  • The zeroeth law says something quite different. It says that if A and B are in thermal equilibrium and B and C are in thermal equilibrium then A and C are in thermal equilibrium. Since it takes as a premise that these things are in thermal equilibrium it isn't deducing that they eventually become in thermal equilibrium. – Timaeus Aug 05 '15 at 17:20
  • Even after your edit you make it a law in the next paragraph. It isn't a law, it's a result of an analysis that only holds when a great deal of conditions hold, it is easy to get objects in contact to have different temperatures even over a long period and you can even have energy flow from hot to cold and make the cold get colder since negative heat capacities are extremely common in gravitationally bound systems, our sun's outer layers will have a negative heat capacity when it gets to a red giant stage. Meeting someone's misconceptions is essential, but condoning it as a law is not – Timaeus Aug 05 '15 at 18:52
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The air has very low thermal conductivity and capacity, in most cases outside, the main contributor to thermal exchange (and thus perception of temperature) is radiation (Stefan's law, every object is radiating light all across the spectrum, with colder bodies giving most of it in infrared, hotter is more visible red (coal embers, hot iron), then yellow, white, bluish white when it gets hotter). In this case, the car is receiving high energy flux from the sun (absorbing most of it), so it's heating very quickly. It's like sitting near the fire - the air between you and the fire is cold, but the fire can still burn you.

Thermodynamically speaking, the car is exposed to the outside air (weakly coupled, slow transfer), the hot ground, the sky (at radiative temperature usually lower than the ambient air, possibly sub zero, if not overcast), and the sun at ~6000K temperature (strong influx, but only from a very specific direction). So you have (sun thermal radiation) + (a very little sky radiation) - (radiative losses of the car) - (conduction, convection through the air). On a sunny day, the temperature gets quite high, before the losses overcome the influx.

Radiative exchange is more important than most people realize. You know how in the winter, you may need a sweater indoors at 25 degrees, but in the summer, a T-shirt at 18 degrees is enough? That's because hot air doesn't help if the walls are cold and don't give that much thermal radiation. Similarly, a sunny day in winter is very cold because instead of relatively warm clouds, you have a "transparent" sky that gives almost no thermal radiation. That's actually the only way the earth is cooling down (and it's substantial, just see how quickly the temperature goes down when the sun sets). On a cold winter night, the surface can cool below the air temperature just because of the radiative losses, and you can actually freeze water on a reflective mirror below the open sky even if the ambient temperature is >0.

So, to conclude... for every object you are calculating the temperature balance for, you have to consider all the objects it's in thermal contact with. That's not just physical contact, every object that is in the line of sight is exchanging the heat via radiation (and in air, that's more important than the direct contact with air). Even in vacuum, where direct exchange is not possible, every object will eventually reach the average temperature of the objects all around it (with objects that take up more angular area contributing more).

That's actually how you can calculate the temperature of the sun just by measuring its size in the sky.

Let's see... the sun has angular diameter on the sky approximately half a degree. That means that out of full $4\pi$ spherical area of the entire sky around the earth, it takes up $\pi (0.5 \pi /180)^2$ (the earth is losing heat all around into $4\pi$, but receiving only from the sun). The Stefan's law states that the thermal flux goes as $T^4$, so the earth temperature will be the average $T^4$ of the sky, so $$T_E^4 = T_S^4 \frac{\pi (0.25\pi/180)^2}{4\pi}=4.76\times 10^{-6} T_S^4$$ If the average temperature of the Earth is $290K$ (give or take), then the surface of the sun is $T_S\approx T_E/(4.76\times 10^{-6})^{1/4} \approx 6200K$. Quite impressive, considering we didn't need any numerical input or natural constants, except for the angular diameter of the sun, which we can measure with a thumb of a stretched out hand.

Maybe I went a bit off topic, but I hope it made things more clear.

orion
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  • " in most cases outside, the main contributor to thermal exchange (and thus perception of temperature) is radiation " Citation? To the best of my understanding convection is far more important as long as you are inside the athmosphere. – Taemyr Jul 21 '15 at 14:18
  • That's actually the only way the earth is cooling down (and it's substantial, just see how quickly the temperature goes down when the sun sets). Deserts are a prime example: may very hot during the day and the temperature drops to sub-zero (°C) during the night. – WoJ Jul 21 '15 at 14:30
  • @Taemyr Convection only works if the atmosphere is out of thermodynamic equilibrium (adiabatic atmosphere, ~1 degree drop per 100m of height). If the ground is really hot, you do get rising columns of air (which eagles use), which dissipate heat into higher atmosphere (possibly starting thunderstorms). But that doesn't change the fact that you are losing $300-400,\rm W/m^2$ with radiation alone! – orion Jul 21 '15 at 14:38
  • @orion Since the inside of the car is warmer than the outside you do not have thermodynamic equilibrium. – Taemyr Jul 21 '15 at 14:40
  • I was talking about the outdoor conditions. For the car, I meant the incoming radiation being the main contributor for heating. However, cooling of a car is also mostly radiation driven. You get $1400W/m^2$ from the sun in perpendicular incidence (probably half in moderate longitudes) and a car at $70^\circ C$ is losing $785 W/m^2$ with its own radiation! Of course, it's also receiving radiation from the ambient - the heated ground, a bit from the sky, but the numbers do suggest that radiation is the biggest contribution both in cooling and in heating. – orion Jul 21 '15 at 14:45
  • Looking at numbers for convective cooling, I'd say about half or less cooling is provided by convection. In windy conditions, convection helps more. The most common numbers for the convective heat transfer coefficients vary ($5-20 W/m^2/K$) so for a $70^\circ$ car, it may be somewhere in the range $200-800W/m^2$. – orion Jul 21 '15 at 14:58
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Perhaps because of the same reasons as the warming of greenhouses? If the windows are uncovered the sunlight increase the energy inside, by isolating the warm air inside the structure so that heat is not lost by convection. [From Wikipedia]

Lehs
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    It's a shame, that this very accurate answer is so downvoted. e.g. Robert Wood had experimentally demonstrated in the beginning of last century, that greenhouses maintain heat inside them not due to trapping IR radiation, but preventing the warm air inside from escaping. Replacing a glass window with a window transparent also in IR does not cause noticeable changes. +1 – LLlAMnYP Jul 20 '15 at 17:01
  • @LLlAMnYP. Thanks, but it was somewhat self-inflicted since I didn't distinguish between the effact in greenhauses and the greenhouse effect in the original post. :) – Lehs Jul 20 '15 at 18:12