What happens to the rate of change of temperature, be it rate of absorbtion or rate of heat loss, when the surrounding pressure of an object rises/drops?
125314
Actually, when u lower the pressure in a chamber of say a liquid..\nIt has to maintaing the Law Of Conservation Of Mass, so the liquid particles evapourate and becomes gas to compensate for the loss..\nthats why you can boil water below 100 deg on the ALPS and thats how they make liq nitrogen..\nNow i dont know whether you are asking in an adiabatic process or not..\nIn an adiabatic process, there is not change in the temperature, the phase change of the particles compensates for the change in pressure...\nBut if its not adiabatic, then basically what u are trying to is crunch many particles into smaller volume or pluck out peices of mass into larger volume..\nIn that case, bonds are created or destroyed. When bonds are created energy is taken in, so temperature goes down...\nWhen bonds are destroyed, energy is released so the temperature increases...\nSo u can synchronize with the absorption and emission rates too.. Its basically exothermic or endothermic depending on pressure...
327828
This is a calorimetry problem! There is an equation for it:       \nq = mc(change in temperature)\n\nq = the heat required \nm= mass\nc= specific heat \n\nAlso, write out the reaction and products.
70571
decrease\nthe greater the surface area of the reactant the more it is exposed to its other reactant, and more of it can react,  thus the reaction rate increases. This is the opposite for a decrease in the surface area.
385559
This seems to be a question about conversion of one type of energy into another.  In this case, gravitational potential energy (Ep) is converted to kinetic energy (Ek), and then to thermal energy (Q).\n\nThe loss of gravitational potential energy when it falls\ndelta Ep = mg(delta h) \nwhere m is the mass of the object, g is the gravitational force constant 9.81 N/kg, and delta h is the change in height,  is equal to its increase in kinetic energy. \n\nI don't really believe that the kinetic energy is entirely converted to thermal energy, but if you assume that it is,\nQ = m (deltaT)c  \nwhere m is the mass of the water being heated, delta T is the change in temperature, and c is the specific heat capacity of water (4.184 J/goC)\n\nSo,    if              delta Ep = Q\n\nthen                      mg(deltah)  = m (deltaT) c  \n   \nand                    (deltaT)  =  g (delta h)/c\n\n\nGood luck !
416534
The question is if you start with gas at 20.0 atm and 20.3 C and heat it to 125 C what will be the pressure?  It is a changing condition problem making use of Charles' Law.  The answer is 27.1 atm.
513350
I will look at it from a different angle.\nConsider the equation of state,\nPV=MRT\nExpansion through a nozzle causes the\npressure to drop and volume to increase \nbut not as compared to drop in pressre. \nThe net effect is that product PV reduces.\nM the mass is not changing. R is the gas \nconstant. So T[temp in absolute] has to drop.
390813
a-decreases\n\nand only about 10% of the energy is actually passed on
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
646059
many materials absorb better at the infrared wavelengths than the shorter wavelengths so the absorbed energy is conserved as heat.  There are materials that absorb shorter wavelengths and convert the energy to heat, such as a microwave oven???  Water absorbs at that frequency (1-2 GHz) and the water molecule resonates.  The vibrational energy is re-emitted as heat due to a Stokes shift.\n\nThe shorter wavelengths are more energetic and more capable of depositing energy that might show up as heat but it depends on the absorption of the material itself.
418611
Your puzzle is explained in part by the thermodynamic behavior of air.  \n\nWhen a 'pocket' of air is heated, its density decreases relative to the surrounding atmosphere, and so it begins to float upward.  As the hot air rises through the troposphere, the pressure around it drops, because there is less air and more available space to fill.  The volume of the hot air expands as a result, but this expansion progressively drains the heat from the air, through a process called 'adiabatic cooling.'  In the end, the air pocket is dispersed, its temperature dropping far below the air temperature at the Earth's surface.  This process happens constantly, all over the atmosphere, and is one factor contributing to the overall 'lapse rate' of the atmosphere (the rate of temperature decline as altitude increases).
268811
both.  the friction, as well as creating its own heat, slows the airflow, which aggrivates the compression along the leading edges.\n\nI'm pretty sure.  don't put this on a physics test and then blame me.
420666
My reference indicates that the speed of sound decreases in a denser substance, rather than increases.  This would be consistent with an increase in velocity in warmer air, because air decreases in density with heat, just as you indicated.
23866
Absorption describes the movement of a gas from the gas phase into a liquid. For example the absorption of acid gases into rain which result in acid rain.\nIt is effected by parameters like the pressure (increased pressure generally tends to give greater absorption i.e. more gas can move into the liquid - think of lemonade when you reduce the pressure by opening the bottle the CO2 undergoes the opposite of absorption, although some Co2 still remains dissolved). It is also effected by the temperature.\nIs is described by Henry's Law which, in simple terms, predicts the amount of a gas which will absorb into a liquid for a given pressure.
409886
Only if E=MC^2\n\n... it follows that if there is a change in the rest-energy of a body, there must be a corresponding change in its rest-mass. For example, if a body is heated, and thereby absorbs a small amount of energy ΔE (as measured in the frame of reference in which the body is at rest), its rest-mass will increase by a very small amount equal to ΔE/c2. This increase is tiny because of the high numerical value of the speed of light. Indeed, for mid-sized objects, such an increase in mass would be too small to measure with even the most accurate balance. For example, if a 1 kg block of gold is heated so that its temperature increases by 10 °C, then, ceteris paribus, its mass would increase by as much as 1.4 × 10-14 kg; a cube of gold of this additional tiny mass would have sides smaller than one one-thousandths of a millimeter. Similarly, if a body emits an amount of energy ΔE, say in the form of light or heat, its rest-mass will decrease by a tiny amount ΔE/c2. In both cases, the important and novel claim made by SR is that the inertial mass of a body can change depending on whether it absorbs or emits energy. This is true, of course, not just for isolated mid-sized objects but for any physical system.
501319
force of air is less than it's weight therefore it rises and it drops.\n\na=b
419059
it *is* Bernoullis principle.  Think of pressure as a form of potential energy and velocity as kinetic energy.  The total energy must remain constant so, if one increases, the other must decrease.\n\n\nDoug
57686
Definition of Rate of change(R):=limit  {[f(x+Δ]-f(x)]/Δ}\n                                  Δ->0\n\nhere x+Δ=1+h and x = 1, therefore, Δ=h;\n\nR = {f(1+h)-f(1)}/h =  {1-(1+h)^6}/h