At what approximate temprature would a diamond be able to evaporate?
479983
At about 1000 degC the surface of a discrete diamond will form a microscopic layer of carbon. The carbon oxidizes to CO2 more readily than does diamond and thus a very hot diamond in oxygen could indeed appear to "evaporate." However a diamond will not support combustion in the conventional sense any more than solid graphite will because the vapor pressure is too low except at ultra-high temperatures. Obviously, porous carbon in the form of charcoal does support embrous deflagration because\nof the high surface area and insulating properties of the \nfuel/air matrix. The same is undoubtedly true of diamod dust, which is somewhat less reactive than graphite.
928091
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.
5132
You of course need to add heat in five separate steps:\n\n1.  Heat the ice from -20C to 0C\n2.  Melt the ice at 0C\n3.  Heat the water from 0C to 100C\n4.  Boil the water at 100C\n5.  Heat the steam to 120C\n\nThe last step (5.) will depend on whether the steam is saturated at an elevated pressure or superheated at one atmosphere which you don't specify.\n\nThe rest is pure calculation using the values for specific heat for ice, water and steam plus the latent heat of freezing (reversed) and the latent heat for evaporation.  Engineers would use a steam table or charts to evalute the steam portion of the problem.
929275
You would need the specific heat of the material.  The rate of cooling would be the total energy loss (E*A) divided by the specific heat.  This will be non-linear, and your value would be good for only the specific temperature given.\n\nWatts = Energy/time; one watt = 1 joule/sec.
321809
Half carat is equal to 100mg and diamond density can be taken as about 3.5g/cc. From this, the size would be approximately equal to a cube of 3mm x 3mm x 3mm or so.
857221
No. The equilibrium constant is thermodynamic. Reaction rate is kinetic. The conversion of diamond to graphite at 1 atmosphere has a high equilibrium constant. Do you see diamonds turning rapidly to graphite?
980695
It depends on how much dry ice you have and what is the temp of the dry ice.
165958
No. Absolute Zero is 459 degrees below zero Fahrenheit. \n\nTemperature is dependent on the rate of oscillation in molecules. At Absolute Zero the oscillations are as low as they can get.
76877
You might be able to make a PM machine based on a vaporizing and recondensing fluid, which is what I'm guessing you're getting at. PM, that is, as long as the necessary temperature differences continue to exist. You would need to vaporize at a low point and condense at a high point, with wheel motion bringing the heavier containers of condensed fluid down to a warmer location for vaporization. It doesn't violate any thermodynamic laws since it takes heat input and exhausts cooler output, just like the water-bobbing woodpecker does. Do it!
655958
The average diamond is approximately 3 billion years old. Some newer some older. Thats about 2/3 the age of the planet.\nDiamonds are not created in the earths crust, but rather deep in the earths mantle at about 1800 degrees F. The diamonds are brought to the surface by volcanic activity.
446916
It differs depending on whether you are speaking of parallell or perpendicular to the surface\nand\nat what temperature it is\n\n8 > 36(10-^6/K)   {@ 0 - 1000°C}\n\nparallel and lower temperatures = lower CTE\n\nCoefficient of expansion per °C parallel to cleavage:\n  0° > 200° C *  8x10^-6 >  9x10^-6\n200° > 500° C * 10x10^-6 > 12x10^-6 \n\nCoefficient of expansion per °C perpendicular to cleavage:   \n 20° > 100° C * 15x10^-6 > 25x10^-6\n100° > 300° C * 15x10^-6 > 25x10^-6\n300° > 600° C * 16x10^-6 > 36x10^-6
960401
You need to know the coefficient of thermal expansion of the brass. This depends on the exact brass alloy that is being used.\n\nAs an example, for brass alloy UNS C35000, the TCE is 20.3 * 10^-6 / °C.\n\nThe basic formula is DL/L = TCE DT where DL means change in length and DT means change in temperature.\n\nYou must expand the sleeve to a diameter greater than the shaft. That means it must expand by at least .02 cm. In general, for an easy slip fit you would want to have about 30 microns of additional clearance - that is .003 cm.\n\nThe fractional expansion is .02/3.195 = 6.3 E-3 for zero-clearance, or .023/3.195 = 7.2 E-3 for 30 microns clearance. \n\nThat corresponds to a temperature difference of 310°C for the zero-clearance case or 355°C for the clearance case.\n\nThese temperature differences are large enough that the assumption of constant TCE will no longer be valid. In particular, you could not achieve this temperature difference by cooling alone, since the projected temperature would be less than absolute zero. \n\nIn reality the amount of interference specified is about a factor of 5 greater than would normally be used.
92881
Maybe this will help (Lexan LS2)\n\nCoefficient of linear thermal expansion (parallel) \n0.7 E-4/°C
464223
'c' is the right answer.\nuse I law of thermo...\nU=Q-W\nU=Q-Q/4\nU=3Q/4\nBut U=n*C*T where n is moles,C is molar heat cap.for const volume.\nLet us assume that the heat absorbed during the process as\nQ=n* A*T\nwhere A can be defined as molar heat capacity for the process.\nso , n*C*T=3*n*A*T/4\nA=4C/3\nFor a diatomic gas C=5R/2(C=no.of degrees of freedom*R/2)\nA=4*5R/3*2\nA=10R/3
149897
Depends on the temperature. At a guess at melting point not that different from solid aluminum.
815534
Diamonds placed in water appear to disappear (other than any flaws in them).  When they are removed from water, they are not wet.
317109
Take table for R134a refrigerant:\n\na) delta s = entropy change\n--inlet for evaporator assuming 50%moisture= interpolate to get S \n@200kpa for R134a\n0       Sg   \n0.5     S\n1       Sf\nget Sg and Sf from table at 200kpa for R134a. \n \n--out for evaporator = Sg at 200Kpa (get from table) since it is sat vap.\n\nthe difference of both Sfg and Sg is the answer.\n\nb)  delta S = Q / T  where Q = 120 KJ and T = -5+273 = 268K\n\nc) add both a) and b) is the total entropy