how much power does the sun transfer to the ground?
478214
Insolation is the incoming solar radiation that reaches a planet and its atmosphere or, by extension, any object exposed to solar rays, such as watts per square meter of Sun-facing cross section, across the entire electromagnetic spectrum; most of that power is in the visible light portion of the spectrum. On Earth's surface, the Sun's rays are attenuated by the atmosphere. The midday insolation on clear days in temperate latitudes may be estimated as 1000 watts per square meter directly facing the Sun. The actual figure varies with the Sun angle at different times of year (because the sunlight must travel through varying distances of air), and with atmospheric haze and cloud cover. Long-term time-averaged insolation in sunny locations is closer to 250 watts per square meter, taking into account the lower insolation in early morning and evening, and the presence of night.\n\nThe insolation into a fixed surface (such as a window or roof) depends on the projected area of the surface perpendicular to the incoming sunlight, not directly on the area of the surface. This is the reason that the polar regions are much colder than equatorial regions on Earth: on an annual average the poles receive less insolation than does the equator, because the Earth's surface at the poles never faces the Sun directly. The projection effect can also be used in architecture to design buildings that are cool in summer and warm in winter, by providing large vertical windows on the equator-facing side of the building (the south face, in the northern hemisphere): this maximizes insolation in the winter months when the Sun is low in the sky, and minimizes it in the summer when the noonday Sun is high in the sky. (The Sun's north/south path through the sky spans over 45 degrees through the year).
1120246
Actually, electrical charges from the ground meet charges from the sky.  We  live in a magnetic field of 219v-dc.   Charges build up in clouds and are attracted to charge harbored within the earth.
1154605
None, but the transfer of energy is more correct.
699167
1. when produced by a generator it is nearly 1\n2. but when it arrives at the use point it can drop to  0.5 if we have a lot of inductive load like motors, transformers etc\n3. In general the lower limit is 0.8 and the power supply companies charge you a penalty for dropping below 0.9 PF\n4, In order to improve your power factor you have to put in capacitor banks to offset the inductive load in your use area.
819964
On the vernal equinox, the sun is directly overhead the Earth's equator at local noon, and we are told that the insolation under those conditions is 1368 W/m^2.\n\nAt latitudes other than the equator, the solar energy per unit area is decreased by a factor of cos(L), where L is the latitude.  This is called the "projection effect".  At 40 degrees latitude, the incoming solar energy per unit area is therefore:\n\n1368 W/m^2 * cos(40 deg) = 1368 W/m^2 * 0.766 = 1048 W/m^2.\n\nI'm assuming the question is asking how long it would take to bring the water to a boil, not how long it would take to completely boil away the water.\n\nIn either case, you are missing a key bit of information necessary to answer the question:  the initial temperature of the water.  Let's assume that temperature is T.  I'll also assume that you are located as sea level, so the boiling point of the water is 100 C.\n\nThe specific heat capacity of water is 4.1855 J/(g*degC), and the density of water is ~1.00 gm/cm^3, so to bring one liter of water from temperature T to 100 C requires:\n\nE = 4.1855 J/(g*degC) * 1000 cm^3 * 1 gm/cm^3 * (100 degC - T) \nE = 4.1855*10^3 J/degC  * (100 degC - T)\n\nWe have a collector with a 1 m^2 surface area, so it supplies 1048 W = 1048 J/sec at 40 degrees latitude, all of which we are to assume goes into heating the water.\n\nWith these assumptions, it would take:\n\nt = [4.1855*10^3 J/degC  * (100 degC - T)]/(1048 J/sec)\n\nseconds to bring the water to a boil.\n\nFor instance, if T = 25 C, t = 299.54 sec ~= 5 minutes to bring the water to a boil.
87409
that the sun gived up
611962
How hot is concentrated light, what is the temp of the sun?\n\nWELL WE HAVE A ENERGY SHORTAGE, ONLY AT SUNSET.\n\n I read about Solar collection in power plants like Solar One.  In these plants the where they make molten salt, for heat transfer for running steam generator. At the reciever of Solar One plant or concentrated light point temperature reachest approximately 1,050 Fahrenheit.\n    \n      THIS LINK IS ABOUT YOUR SUN AND THE TEMPS ARE HOT.\n         SO HOW MUCH LIGHT CAN WE CONDENTRATE MORE  \n           DEVELOPMENT IN SOLAR SAILS\n\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html \n\n Earth's Sun is a medium-sized star which lies on the main sequence with 90% of the known stars. It has a effective surface temperature is 5780 K, putting it in spectral class G2. Its mass is 1.989 x 1030 kg and its mean radius is 6.96 x 108 meters. The mass of the sun is over 99.8% of the mass of the entire known solar system, leading de Pater and Lissauer to refer lightly to the solar system as "the Sun plus some debris". \n\nThe sun radiates energy at the rate of 3.85 x1026 watts. Just outside the earth's atmosphere solar energy is received, assuming normal incidence, at the rate of 1340 watts per square meter. \n \n\nThe orbit of Earth ranges from 1.47 to 1.52 x 1011 meters from the Sun. The average light travel time to the earth is 8.3 minutes. \n\nThe radius of the sun at 696,000 km is 109 times the Earth's radius. Its surface gravity is 274 m/s2 or 28.0 times that of the Earth. Its mean density is 1410 kg/m3 or 0.255 times the mean density of Earth. \n\nThe composition of the sun is 71% hydrogen, 27.1% helium and less than 2% of all other elements. \n\nThe center temperature is modeled to be 15.5 million K. The Sun is fueled by the proton cycle of nuclear fusion.\n\nEscape velocity = 618 km/s\n\nSO
156733
Sun has a nuclear reaction (fission of Hydrogen gas)to radiate energy. Theoratically as much as Hydrogen is being used in the process of energy production the net quantity of hydrogen is being reduced and so net amount of reacting atoms is decreasing. Thus energy of sun is being reduced. However, the rate of decrease in the active mass is so slow that it can not be measured in course of few centuries.\n\nThe much referred worming of air or global worming, as replied here by one other, is due to increasing pollution of air or green effect.
808740
It depends on how much you use the electricity.
604131
Extrapolate the formula, using the surface temoerature, and instead of the earth distance, substitute the distance you want to caluclate the temperature at.  \n\n\nMeasuring the Sun's Temperature\nOne method, called Wien's law, uses the wavelength of the peak emission, wavelengthpeak, in the Sun's continuous spectrum. The temperature in Kelvin = 2.9 × 106 nanometers/wavelengthpeak.\n\nAnother method uses the flux of energy reaching the Earth and the inverse square law. Recall from the Stellar Properties chapter that the flux is the amount of energy passing through a unit area (e.g., 1 meter2) every second. From the Inverse Square Law of Light Brightness, you find that the solar flux at the Earth's distance = the Sun's surface flux × (Sun's radius/Earth's distance)2 = 1380 Watts/meter2. Since the Sun's photosphere is approximately a thermal radiator, the flux of energy at its surface = sigma × (the Sun's surface temperature)4, where sigma is the Stefan-Boltzmann constant. Rearranging the equation, the photosphere's temperature = [(solar flux at Earth)/sigma) × (Earth distance/Sun's radius)2]1/4.\n\nThese two methods give a rough temperature for the Sun of about 5800 K. The upper layers of the photosphere are cooler and less dense than the deeper layters, so you see absorption lines in the solar spectrum. Which element absorption lines are present and their strength depends sensitively on the temperature. You can use the absorption line strengths as an accurate temperature probe to measure a temperature of about 5840 K. \n\nRead the link, you'll have to scroll down to come up to the stuff I have cut and paste above.\nst\nHope you find this useful.
662957
Many people are often confused about power and energy.\nLet's start from the beginning, with some basic equations:\n\nEnergy = Force * distance\n\nFor a mass m moving at a speed v, the energy E is:\n\nE = 0.5 * m * v * v\n\nEnergy can be measured in Joules (J) or British Thermal units (BTU). 1 BTU is equal to roughly 1055 Joules. If a 2100 pound car is travelling at 60miles/hour, the energy is about 342600 Joules or 325 BTUs. This is the same amount of energy required to accelerate the car from 0 to 60 miles/hour on level pavement. This is independent of acceleration, it doesn't matter if it takes 6 seconds or 60 seconds, it's the same amount of energy.Energy is always conserved, meaning it can't created or destroyed, it can\nonly change form.\n\nPower = rate of energy\n\nPower can be measured in Watts (W) or Horsepower (HP). 1 HP is the same as 746 Watts.\n\nPower, unlike energy, is not always conserved. If we have a system where energy is flowing at a constant rate, then power is conserved, meaning the power in is equal to the power out. For standard cars, power from the engine is delivered to the wheels through a transmission. There are some losses due to the transmission, air conditioner, alternator, and other systems, but this is not considered a significant loss. If the air conditioner is turned off, your headlights are off, and you have a manual\ntransmission, the losses will be not more than a few percent. Automatic transmissions only increase losses.\n\nThe gear ratio is a transmission does not change power output. Remember,power is conserved in this case. There might be some slight change in power loss for different gear ratios, but for the most part, it's insignificant.\n\nSince we know power is the rate of energy, we can divide the energy calculated above by the time required to produce the energy. If it takes your car 6 seconds to go from 0 to 60 miles/hour, then the power is about 57000 Watts or 76.6 HP. Keep in mind, this is the average power, it doesn't\ntell us the power at any time during the acceleration. We would need detailed speed measurements from start to finish to calculate that.\n\nGiven these numbers, I would guess the engine is capable of producing 150 HP or more. How did I come up with that number? Internal combustion engines, like gasoline, Diesel, and even turbines, do not have a flat power curve. The power output is zero at zero RPM and increases more or less\nlinearly until some peak. In stock gasoline engines, the peak power is at 5000-6000 RPM or so. As a rule of thumb, the peak is generally 1000 RPM below red line. So if you could run your engine at 1000 RPM below red line,\nthen you could achieve near peak performance. If watch your tachometer, I would guess that the RPM averages much less than that, maybe even half.\nThat's why you are getting only half of the power the engine is capable of producing.\n\nRPM, power and torque are all related. Torque is simply power divided by RPM. The power out of a transmission is the same as the power in (minus small losses), but the RPM is different depending on the gear ratio. If the output shaft spins at half the speed of the input, then the output torque is double that of the input. RPM doesn't really matter in the end as far as power is concerned, remember, power is conserved. Changing gear ratios does not change the power, it only affects torque and RPM. It's easier to start with calculate power and work out RPM and torque if you need it. You can do it the other way, but it's easy to make mistakes.\n\nThe reason most cars have a transmission at all is because gasoline engines cannot produce torque over a wide range of RPM, particularly at zero RPM.Diesel electric locomotives and subways do not have transmissions with variable gear ratios because they use electric traction motors instead of\ntransmissions. Electric motors can produce torque over a wide range, even at zero RPM. This is known as the stall torque
604274
It takes light 8 minutes (plus or minus seconds) to travel from the Sun to the Earth. So if the sun 'Died' this second, it will take 8 minutes till we know about it, i.e. the lights go out and the heating is turned off.\n\nHopefully we will still have the stars and moon to merrily sooth us into our freezing tombs.
241725
The ultimate source of energy in all. living systems is the sun
676858
150 million kms divided by 300000 km/sec\n8 minutes 20 seconds.\n\nHowever... now comes the fun part.\n\nFusion, the energy source of the sun, takes places only in the core. And the sun is so dense, that a photon created down there, takes a lot of time to come out to the surface. It collides, gets absorbed, is re-emitted, takes a day of to dance.\n\nIn all a photon takes a million year or so to surface.\nAnd million years plus 8 minutes to arrive at the earth.
25250
I found what you may be looking for at the website below.  Here it is...\n\nFirst it should be known that the Earth's orbit around the Sun is elliptical, not perfectly circular, so the Earth-Sun distance is changing as we speak just from the Earth traveling in its orbit around the Sun.  \n\nIs the orbit itself changing? Well, there are some long-period oscillations, but those are very small, and don't imply that we're systematically moving towards or away from the Sun. \n\nThere is an effect which is making us move very slowly away from the Sun. That is the tidal interaction between the Sun and the Earth. This slows down the rotation of the Sun, and pushes the Earth farther away from the Sun. You can read more about tides, as they relate to the Earth-Moon system. The principle for the Sun-Earth system should be the same. But how big of an effect is this? It turns out that the yearly increase in the distance between the Earth and the Sun from this effect is only about one micrometer (a millionth of a meter, or a ten thousandth of a centimeter). So this is a *very* tiny effect. \n\nThere is another effect which is also small, but somewhat bigger than the tidal effect. The Sun is powered by nuclear fusion, which means the Sun is continuously transforming a small part of its mass into energy. As the mass of the Sun goes down, our orbit gets proportionally bigger. However, over the entire main sequence lifetime of the Sun (about 10 billion years), the Sun will only lose about 0.1% of its mass, which means that the Earth should move out by just ~150,000 km (small compared to the total Earth-Sun distance of ~150,000,000 km). If we assume that the Sun's rate of nuclear fusion today is the same as the average rate over those 10 billion years (a bold assumption, but it should give us a rough idea of the answer), then we're moving away from the Sun at the rate of ~1.5 cm (less than an inch) a year. I probably don't even need to mention that this is so small that we don't have to worry about freezing.\n\n:)\nPeace!
865349
Energy in the form of photons is generated in the sun's core. Photons like to take off at the speed of light, so they do. But they move only 1micron before running into(being absorbed by) a gas molecule. That's because the gas atoms at the center of the sun has a pressure of 250 billion times the earth atmosphere's sea level pressure. That is why gas atoms are so hard for photons to miss. But the gas atom gets so excited with the extra energy it carries from that hot photon that it lets go of another photon, that promptly runs into another gas atom. This emitting & absorption occurs 100 million trillion times PER YEAR.  You'd think all this elbowing going on that it would get to the sun's surface fairly quickly. But this radiative process takes from 30,000 to 200,000 years to make it only about 55% of the way to the sun's surface. But once it makes it to this higher convective layer of the sun(like the earth atmosphere's lower convective atmosphere), then the energy is transported fairly quickly to the sun's surface.
59746
the sun weighs approximatly 2.0 * 10^30 kg\n\nto abisdad2004's answer i believe the site has a typo, they probably meant to the 30th power