what is the tempeture in the troposphere?
152908
The average temperature of the troposphere (0 to 6 miles above Earth) is 59-degrees Fahrenheit.
284973
-213.8 °F (brrrr).\n\nTemperature is proportional to the kinetic energy of air molecules in motion. At absolute 0, there would be no molecular motion, which is -459.67 °F. Thus 32 °F is 491.67 °F above absoltue zero.  If the molecules have half the energy as that, I suppose it could be considered twice as cold (half as warm = twice as cold?). 491.67/2 = 245.8 °F above absolute 0 and below 32 °F. So 32 - 245.8 = -213.8 :)\n\nUsing the absolute (Kelvin) temperature scale simplifies this calculation because absolute 0 is 0 K.
144701
by defn:\ntemperature is the degree of hotness or coldness of a body whereas heat is the energy in transit.
225559
54.36 K\n(-218.79 °C, -361.82 °F)
591984
It gets colder as you go up through the troposphere because the air is "thinner".\n\nThere is less atmosphere here due to less gravity.\n\nB/c there is less air, there will be less heat. Air is an insulator that holds heat. Less air = less heat.\n\n:)
115476
The mesosphere (from the Greek words mesos = middle and sphaira = ball) is the layer of the Earth's atmosphere that is directly above the stratosphere and directly below the thermosphere. The mesosphere is located about 50-80/85km above Earth's surface. Within this layer, temperature decreases with increasing altitude. The main dynamical features in this region are the atmospheric tides which are driven by momentum propogating upwards from the lower atmosphere and extend into the lower thermosphere
266705
for every 167 meter increase in the first layer of the atmosphere there is one degree Celsius decrease in temperature.If you get the temperature of the gas,in this case atmosphere, you can calculate the pressure by using the ideal gas law equation.
1031939
temperature decreases with increase in altitude
806362
Atmospheric characteristics\nAtmospheric pressure 140 kPa\nHydrogen> 93%\nHelium >5%\nMethane 0.2%\nWater vapor 0.1%\nAmmonia 0.01%\nEthane 0.0005%\nPhosphine 0.0001%
689681
(m)\n\nAt a temperature of 15 degrees Celsius, the speed of sound in air at sea level is approximately 341 m/s, and is proportional to the square root of absolute temperature; symbol, c.
129713
MeghnarMajhi is right.\n\nTemperature <==> Heat
1124152
tropo = 1 : turn : turning : change <troposphere>\nmeso = 1 : mid : in the middle <mesocarp>\nstrato = is related to layers, or stratification\nthermo = 1 : heat <thermostat>\nexo = 1 : outside <exogamy> : outer <exoskeleton>
602468
Absolute Zero or around 20 K.\n\nMelting point 14.01 K  (−259.14 °C, −434.45 °F) \nBoiling point 20.28 K  (−252.87 °C, −423.17 °F)
899540
91.5°East\n\nhttp://en.wikipedia.org/wiki/Measat-1
456439
Air is heated & rises at the equator. The air is forced to about 30 degrees north & south of the equator at high altitude. It starts to cool at these latitudes and starts to descend. As it descends it is heated due to decreasing antitude and increasing atmospheric pressure. This causes the moisture content in the air to decrease resurlting in typically cloudless skies and little rain. This is called a 'Hadley Cell'.\n\nThese latitudes were called the horse latitudes becuase in the days of sailing ships which were often becalmed in these latitudes, sailors were forced to kill livestock on board ships to conserve water and provide food.
105710
To expand on Valentin's answer:\n\nFirst, realize that most atmospheric heating comes from the surface of the earth; the atmosphere itself is transparent to most frequencies of light energy coming from the Sun.  It's when that radiation hits the surface of the Earth and gets converted to infrared, i.e. heat, radiation that the atmosphere gets heated.\n\nNow, the "physics problem":  Pressure, density and temperature are all related by gas laws.  As the warm air rises, it expands, without taking on heat from its environment, and thus cools.  \n\nNext, add the "natural phenomenon".  Higher altitude air is generally drier than lower altitude air; water vapor helps lock in the heat radiating from the surface; hence higher altitude air locks in less heat than lower altitude air.  And, it's true that high altitude winds result in the air absorbing less heat from the surface.
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