The object is wooden, about 40 meters long and 11 meters high and is burried under 1 to 1.5 meters of soil and vegetation.  The climate is tropical and it would be in the summer time. It may have standing water within the object to 50% of it's available interior. How much would the interior air temp. differ from the ambient air temp. surrounding the object? What instrument would best mesure this difference?
127429
I think this guy is trying to grow something hydroponically and doesn't want anyone else to detect it.  Know Your Enemy.  At any rate, since there is so much water, there probably will be an increase in vegetation around the object in question.  As far as the difference in temperature, since water is a great heat sink, the water would probably end up being about the same temperature as the surrounding soil.  When you get more than 3 ft deep (Except in possibly extreme climates) the soil temperature stays about the same year round.  The sun just can't warm and the air just can't cool the soil fast enough for the underlying substrates to be affected.  This is why when utilities (gas, electric, water, etc) are buried, there is a minimum depth for the conduit to be below the ground so that changes in temperature will not affect the performance and/or integrity of the circuit, in addition to the fact that the average guy can't just come along with a shovel and damage it.  One thing that may affect the temperature of the buried object is how well the soil it was buried in was placed around the object (i.e. the density of the soil placed).  If the area was just backfilled with no attempt to pack the soil, the soil would have large pockets of air in it which would allow it to not only retain the heat it has, but also settle relative to the surrounding natural soil.\n\nI hope this helps.
547453
Let's say a mass of air travels across Nevada and Utah and arrives in Colorado.  This is what generally happens anyway because of the prevailing westerlies, as you know.\n\nThis mass of air has moisture evaporated in it, basically from water that evaporated from the Pacific Ocean.  (In simplistic terms...) the air may have a temperature of 55 degrees, and a dewpoint of 45 degrees.  i.e., if it were to be cooled to 45 degrees, the relative humidity would become 100%, and it could not hold any more water molecules.\n\nIt is forced to rise as the prevailing westerlies force it up and over the Rockies.\n\nI'll ignore the question about why it is cooler at higher altitudes/elevations, because answering that would make my answer way too long, but suffice to say that the (dry adiabatic) lapse rate is 9.8 degrees celsius per 1km, or roughly 6 degrees per 1000 feet.  A mass of 55 degree air that crosses the western USA at around 4000 feet elevation, then gets pushed up to 12,000 feet of the Continental Divide expands and cools by about 48 degrees.  (8000 feet times 6 degrees per 1000 feet)  Again, I'll warn that this is tremendously simplified.  As it crosses the Divide, it is cooled to about 7 degrees F.\n\nNow, since that air has a dewpoint of 45, what must happen?  Condensation.  The cooling condenses out the moisture in the form of snow (and also rain at lower elevations), and the air is "dried out" to a new dewpoint of that same 7 degrees. \n\nThe air travels on down to Denver and points eastward  (This is also why it is relatively dry on the High Plains, and few trees are seen there, but rather it is an ocean of prairie grasses, which can survive with less rain than trees need.), but the snow remains on top of the mountains!\n\nYou might also try searching "orographic lifting" if you want to learn more.
498335
Generally the most important is temperature. the Human body is most comfortable between 18 - 21 degrees (around this). This varies slightly depending on the temperature outside. Humidity. Another is light. Diffused light from the sun is more comfortable than direct sunlight as we do not like glare. \n\nFactors like these vary depending on what is happening within the building e.g a sports facility would need to be cooler because people get hotter exercising and it becomes more humid because of the moisture created by people.\n\nThese factors can be measured using items such as a thermometer, humidity gauge (don't know the name) and a light meter. \n\nThis is all I know about the subject.\nI hope it helps.
373237
This question is not very difficult, but you have to be careful with the unit.\n\nFirst, to solve this question, you have to make the assumption that there is no energy lost to the surrounding. So, this reduces to a two system question: one system releases energy while the other one absorbs energy. In this particular case, the energy released by one system equals to the energy absorbs by another system.\n\nTo picture this, consider the following, one system goes through the following steps:\n\n[100 degC H2O (g)] --- (step 1) ---> [100 degC H2O (l)] --- (step 2) ----> [79 deg C H2O (l)]\n\nThe other system goes through:\n\n[260g, 19 degC H2O (l)] --- (step 3) ----> [260g, 79 degC H2O (l)]\n\nIt is obvious that both the energy involved in step 1 and 2 is negative (release energy) while the energy involved in step 3 is positive (absorb energy). We will now solve this question backwards. Starts from finding out the energy required by step 3, then find out the mass of steam that is needed to do the job.\n\nEnergy absorbed during step 3:\nE3 = m * c * (delta T) = 260 * 4.18 * (79-19) = 65208 J\n\nEnergy released in step 1:\n(you have to first convert heat of vaporization to unit of J/g: 40.7*1000/18 = 2261.111 J/g)\nE1 = m' * l = m' * 2261.111 = 2261.111m'\n(since we don't know how much steam is needed, let's just leave it as a variable)\n\nEnergy released in step 2:\n(the temperature of the same mass as above will be lowered to 79 degC)\nE2 = m' * 4.18 * (100-79) = 87.78m'\n\nTotal energy released = (2261.111 + 87.78)m' = 2348.89m'\n\nEnergy released = energy absorbed\n2348.89m' = 65208\nm' = 27.76 g
229913
Inside or outside volume?\n\nInside:  Can't use Fluid => Gas or liquid.  Use solid like sand to fill the vase.  Takeout the sand and measure the volume either by weight/density or use another standard measuring device.\n\nOutside: Trace the curve, find out how many continuous sections are there that are connected at discontinuous boundaries, find volumes of each (use cylinder or cone volume formula) and add up the volumes.
527641
Depends of if the body temperature is the same as the surrounding, for example a snake that has been sunbathing for a few hours before dark will remain warmer than its surroundings for some time, so thermal immagery devices could have limited use. I'd say a flashlight or headlamp would probably be just as much use in most cases tough.
205604
The sun heats the ground. \nThe warm ground in turn heats the air above. (In the above diagram, the warm ground and air are shaded in red.) \nThe warm air at the ground is lighter and less dense than the surrounding air above, so it begins to rise. \nThe rising air takes on the shape of a round "parcel." \nSeveral things happen as the parcel rises up through the atmosphere:\nThe higher up into the atmosphere you go, the lower the pressure. As the parcel of air rises, it experiences this drop in pressure. As the pressure on the air parcel drops, the parcel expands and increases in size. (In the diagram I drew the rising parcel getting bigger as it rises.) \nAnytime air expands due to a drop in pressure, it also gets cooler. So, as the parcel rises and expands, it also cools. (In the diagram, the green and blue colors indicate the parcel getting cooler.) The parcel cools at a rate of approximately 10 degrees C per 1000 meters of rise (or 5.5 degrees F per 1000 feet of rise.) This is called the dry adiabatic lapse rate. \nWarm air is able to hold larger amounts of water vapor (water in gas form) than cool air. The cooling parcel is no longer able to hold all of the water vapor it was able to hold when it was warm. The excess water vapor begins to condense out of the air parcel in the form of liquid cloud droplets. When water vapor begins to condense into liquid droplets (moisture), this means the air parcel is saturated and has reached its dewpoint temperature. The higher the parcel rises, the more it expands, the cooler it gets, and the greater the amount of moisture that condenses out of it. \nThe rising, expanding, cooling air parcel finally rises so high that the surrounding atmospheric air is the same temperature as the cooled parcel. This is called equilibrium temperature. When equilibrium is reached, the parcel loses momentum and stops rising. \nBy the time the air parcel has reached equilibrium temperature, enough moisture has condensed out of it to form a visible cloud! \nCloud heights and shapes vary according to the atmospheric conditions. On a hot day, clouds will tend to form higher, because the rising air parcels must rise higher in order to cool enough to reach dewpoint. When air parcels continue to rise after dewpoint has been reached and condensation has occurred, the cloud becomes very tall. This is how cumulus congestus and cumulonimbus clouds form. If air parcels only rise a short distance over a large geographical area after reaching dewpoint, layer clouds (such as stratus) will form.
511821
In general the temperature of the water has very little impact on the temperature reduction due to evaporative cooling.\n\nAs to how cool you can get the air. If you have sufficient contact between the air and water, you should be able to approach the wet bulb temperature of the air but since the contact is not 100% efficient you will never reach the wet bulb temperature.\n\nI don't have a chart handy but you should be able to determine the wet bulb temperature that corresponds to 23% humidity.
370898
V1/T1 = V2/T2, where T is measured in Kelvins!!!\n\nV2 = V1*T2/T1\n\nV1 = 69.0 mL\nT1 = 49.0°C = 322.2K\nT2 = 89.0 °C = 362.2K\n\nNow you can do the arithmetic! (Hint, the balloon will be larger!)
616896
Afjas, I would answer if I knew, but my thing is LONG distance time table comparison. I would guess that you could run it indoors about 5 seconds faster due to no wind resistance, but if it hotter inside than outside, you can take away that advantage.
200260
somewhere around room temp\n\nin a freezer
15915
The formula for calculating specific heat is q=mC(T2-T1). Where q=heat exchange, m=mass, C=specific heat, T2=final temp, and T1=initial temp.\n\nSince you have two materials inside your container, your formula would be q=[mC(T2-T1)]for the water + [mC(T2-T1)] for the metal.\n\nThe container is insulated so there is no heat exchange between the outside environment and the materials in the container. So q=0.\n\nThe formula reduces to 0=[mC(T2-T1)]for the water + [mC(T2-T1)] for the metal. \n\nOr [mC(T2-T1)]for the water = -[mC(T2-T1)] for the metal\n\nC of metal= [mC(T2-T1)]for the water/[-m(T2-T1)]for the metal.\n\nC of metal= [(185g)(4.184 J/gK)(303.75K-298.15K)]/\n            [-(36g)(303.75K-451.15K)]\n\nC of metal = 7775.313J=7.775kJ
410473
Use the Gay-Lussac gas law: V/T is constant. Note that T here is the temperature in Kelvin, not in Celsius. Luckily 27 C = 300 K
650830
Water acts as the medium in many processes. 25 deg C is the room temp.
406205
Its maths again!\n\ndiffence in temp is 2.75 C\n\nWeight of silver is 4.37 g so your anwer is\n\n4.37 g x 2.75 C x 0.24 J/gc
108946
Have each page work on the same princable as the bi-metallic thermometer.  Try to find two materials the expand differently than eachother because of heat, or you could use humidity to expand one material more than the other. You could probably find some material that contracts (or expands) when an electrical current passes through it. place a strip of this along the top and bottom of each page. Thats all i can think of right now.\nCool question by the way.
573134
At that depth the temperature is likely to be close to the annual average temperature. Your idea is excellent and is commonly applied commercially in the form of geoexchange pumps. \n\nOne reason this works, besides the fact that shallow ground is going to be near annual average temperature, is that there is an observed delay in the change of ground temperature at some shallow depth that will be about 6 months out of phase with the surface temperature. So in the summer when you need it most, there will be a cooler zone in the ground beneath you that is still affected by last winter's cooling. \n\nThe SMU Geothermal Lab has a lot of data relative to geothermal gradient. However, geothermal gradient is the 'rate of change' in temperature and not the actual temperature data, so I went through their spreadsheet data and tried to find a well temperature that would be relevant to you. Few wells have temperature readings taken at such a shallow depth as 60 meters. Here is their site, which you may find interesting:\nhttp://www.smu.edu/geothermal/\n\nWhat you are really interested in is called a geothermal heat pump. Here is a page with links to lots of information on geothermal heat pumps:\nhttp://www.smu.edu/geothermal/GPW_TX/GPW_TX_HeatPumps.htm\n\nTake a look at some of this information:\nhttp://www.geoexchange.org/publications/factsheets.htm\nGeothermal heat pumps have been found to be one of the most efficient methods for heating and cooling buildings, with an average of 36% more efficiency for heating and 43% more efficiency in cooling that gas heat pumps. There are over a million installations of geoexchange pumps in the US at present. Here is a diagram that shows how it works for cooling:\nhttp://www.geoexchange.org/images/Home%20GX%20smmr.jpg\n\nHere is another good writeup by an engineer on geoexchange:\nhttp://www.greenbuilder.com/sourcebook/groundsource/\n\nHere is the EPA's site on geoexchange:\nhttp://www.epa.gov/globalwarming/greenhouse/greenhouse9/comfort.html