Formally derive the temperature of the Earth for 3 assumptions:?
829976
You'll need this number:\n\n1.740×10^17 W.\n\nThis is the solar constant, the amount of energy received at the earth per second per area.\n\nAnd this number:\n\nsigma=5.67x10^-7 J/s/m^2/K^4\n\nThis is the Stefan Boltzmann constant which goes into the (surprise) Stefan Boltzmann law:\n\nJ=e*sigma*T^4\n\nSo, J is the energy received by the body per second per area, and T is it is temperature. The e stands for the efficiency of the body which is the ratio between the power it absorbs and then reemits.\n\nSo, you have an albedo which is an indication of how much light an object reflects from its surface versus how much it absorbs. With these numbers, you should be able to piece together your information with the equation above to get an answer. Let me know if you have further questions.
768228
a. V ice = pi(0.12^2-0.10^2)*3 = 0.041 m^3\nb. V ice = pi[(0.10 + x)^2 - 0.10^2]*3 = pi(0.10^2 + 0.20x + x^2 - 0.10^2)*3 = 3pi(0.20x + x^2)\ndV/dt = pi[0.60 + 6(0.02)]dx/dt = 0.000002pi\n= [0.60 + 6(0.02)]dx/dt = 0.000002\ndx/dt = 0.000002/[.60 + 6(.02)]
151603
a) set TC = 0 since celsius is on the x-axis and you find the y intercept by setting whatever is on the x-axis equal to zero.\nb) 32 degrees farenheit.\nc) m = 5/9  y-intercept = -5/9*32 = -17.8 TC
976898
P1V1/T1 = P2V2/T2\nP1 = 1 atmosphere = 101325 N / (m^2)\nV1 = (4*10^3 cm^3)(1 m^3/10^6 cm^3) = 4*10^-3 m^3\nP2 = P1 + h*2.00103/0.01\nV2 = V1 + 0.01h\nT1 = 20°C = 293.15°K\nT2 = 260°C + 273.15 = 533.15°K\nA = 0.01 m^2\nP1V1/T1 = (P1 + hk)(V1 + Ah)/T2\nP1V1/T1 = P1V1/T2 + ((kV1 + AP1)h + kAh^2)/T2\nP1V1(T2 - T1)/T1 = (kV1 + AP1)h + kAh^2 \nkAh^2 + (kV1 + AP1)h - P1V1(T2 - T1)/T1\nh^2 + ((kV1 + AP1)/kA)h - P1V1(T2 - T1)/(kAT1)\n((kV1 + AP1)/kA) = ((2.00103*4*10^-3 + 0.01*101325))/(2.00103*0.01)\n((kV1 + AP1)/kA) = 50636.82\nP1V1(T2 - T1)/(kAT1) = (101325)(4*10^-3)(533.15 - 293.15)/(2.00103*0.01*293.15)\nP1V1(T2 - T1)/(kAT1) = 16582.28\nh = (- 50636.82 ± √(50636.82^2 + 4*16582.28))/2\nh = - 25318.41 ± √(25318.41^2 + 16582.28)\nh = - 25318.41 ± 25318.74\nh = 0.34 m = 340 cm\n\nb)\nP2 = 101,325 + 0.34*2.00103/0.01\nP2 = 101,393 N/m^2
521925
You find it in the normal coordinates way; i.e.\n((x1+x2)/2, (y1+y2)/2)\n\nThe curvature of our Earth does not affect it as the coordinates we are using is 2-dimensional.
265928
1.27 X 1022 KG (0.0021 Earths)
964851
This is an old math riddle.  Mathemeticians and physisists like to refer to absolute zero (Kelvin scale, with no negative numbers), instead of Fahrenheit or Celsius (which are relative to the freezing and boiling points of water at sea-level).  \n\nTechnically, in a relative scale it makes no sense to speak of doubling or halving the temperature.  (If it is 0 degrees (F or C), then doubling or halving would still be 0, which makes no sense.)  \n\nA less scientistic but more human approach is to postulate a base temperature that most people find comfortable.  Let's say it's 70F (21C, 294 Kelvin, 529 Rankine, 17 Reamur).  If today is "cold", it means less than 70F.  E.g., maybe today was 50F.  So it was 20 degrees (F) colder than is comfortable (70 - 50 = 20).  If tomorrow will be "twice as cold," it will be 70 - [2 x (70 - 50)] = 70 - (2 x 20) =  30F.  Quite cold!
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Fahrenheit  to Celsius °C = (°F – 32) / 1.8\nCelsius to Fahrenheit °F = °C × 1.8 + 32
362695
The solar constant for Earth is sc = 1360 W / m2 = 1.36 GW / km2 [1]. Assuming that the radius of the Earth is RE = 6,366 km then the total rate of energy flow (power) on Earth is equal to: \n\n\nPE = sc    RE2  1.7  108 GW (1) \n\nSome of this energy accumulates in the Biosphere, but most of it is radiated back to space in the form of electromagnetic radiation. \n\nThe temperature of the Sun's surface is TS  5800 K. Assuming that the Sun is a black body at temperature TS = 5800 K and that all of the energy impinging on Earth from Sun is radiated back into space in the form of black body radiation it can be shown that the Earth black body temperature TE is [1] : \n\n\nTE = (0.25) 1 / 4  (RS / RSE) 1 / 2  TS  280 K (or  7oC) (2) \n\nwhere: RS  7  108 m - radius of the Sun, RSE  1.5  1011 m - average Sun-Earth distance \n\nThe Sun is the major energy source for Earth. Geothermal energy is comparatively minor, since ice caps formed on the Earth's Poles where the average intensity of the Sun's radiation is low.
1136926
A simple solution is to mulitply the celsius temperature by 2 and then add 31 (32 - 1). It's not the exact science, but It will give you the right answer 97% of the time.
604439
Minus 127 Fahrenheit in Antarctica.
443794
30 degrees or pi/6 radians
45840
Could be. There is so far no way to know the temperature of the earth's core. We know from drilling that the temperature increases about 1 degree Fahrenheit for every 60 feet in depth. Of course, there is considerable variation in this from region to region, since it depends on the thickness of the crust, volcanic activity, ocean depth, etc. Anyway if you extrapolate that towards the core, it would give you an answer much greater than 12,000 degrees F. Most sources estimate about 7500 Kelvin or 13,040 F. We can attempt to model the core to calculate the temperature but since we don't know for sure what it is made of, and cannot reproduce the extreme pressure, it is hard to really know the temperature.
266364
It's 4000 degrees C at the center of the Earth. That'd pretty much do the trick.
907651
Just add first two and absolute value of third... answer 2 degree C.\nIf the values are the temperature level after the change... net change is from level 1.1 to minus 1.9 ...so net cgange is 1.1+1.9=3 deg C.\n Exact Wording of question is important.
454506
100 degrees celcius\n212 degrees Fahrenheit
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Heath Earth Science. Heath, 1994: 22. "The inner core is probably no hotter than 7000 degrees Celsius." 7300 K \nThe Worldbook Encyclopedia. Chicago: Worldbook, 1999: 20-21. "The temperature at the core maybe as high as 7000 degrees Celsius." 7300 K \nDe Bremaecker, Jean-Claude. "Temperature in the core." Geophysics of the Earth's Interior. Canada, 1985: 296-297. "The temperature at the inner core boundary is probably between 3400 and 5700 degrees Celsius." 3700 - 6000 K \nBeiser, Arthur. The Earth. US: Time Life, 1969: 36. "It's temperature is somewhere between 2000 and 4500 degrees Celsius." 2300 - 4800 K \nKubala, Bizy & Mahan Rao. Earth's Core Temperature. Byrdand Black, 1996. "Earth's core temperature is estimated at around 5000 to 7000 degrees Celsius." 5300 - 7300 K