How  much pressure does it take to increase the temperture of water by 10 degrees if the water is 75 degrees?
409423
You have to specify the energy migration profile. Either adiabatic (zero energy transfer) or iso-volumetric (a lot of energy entering the system) or something in between.\n\nAlso, you have to specify the thermal units (Celsius/Fahrenheit/Kelvin/Rankin).
860053
measure the cold water temp and take a large amout of cold water than hot water and mix them .\nuse the realation mhot (Thot - T2) = mcold(T2 - Tcold)\nwhere:\nmhot = amount of hot water , Thot = hot water temp\nmcold = amount of cold water , Tcold = cold water temp.\nT2 = the temp. of water after mix.\n(considering the specific heat of water constant at this range of temp.)
797871
raising the temperature. like with a saturated solution of salt-water, if you reise the temperature the water will take in more salt. it's the same with other thing too.
845517
One way is to use the Clausius-Clayperon Equation.  See the link below and use the boiling point of water for one of the pressure and temperature pairs.  \n\nP.S.  It's equation 22 in the linked document.
186344
At 20 degrees C, the vapor pressure is 17.54 mmHg.  At 25 degrees C, it's 23.76 mmHg.  You'd get close if you interpolate between these two points, even though the curve, strictly speaking, is not linear.  \n\nOr, if you're ambitious, use the Antoine equation:  \n\nAntoine Equation Parameters\n\nlog10(P) = A − (B / (T + C))\n\n    P = vapor pressure (bar)\n    T = temperature (K)\n\nFor temps in the range 273. - 303. Kelvin, A = 5.40221, B = 1838.675 and C = -31.737.
904003
Temperature rise of water = 2 deg C = 275 K\nq=mc(change in temperature)\n  = (100)(4.18)(275)\n  =114950J\n  = 114.95 kJ\nHmm. Either I'm wrong or the question is incomplete. Did the question perhaps give more information on the reaction?
360729
It doesn't.  The t(exp 4)(temperature difference to the 4th power) law works here.  The higher temperature water will cool faster than the cooler water, however, when it reaches the temperature that the cooler water was in in the first place, it will cool at the same rate the cooler water cooled with.
317634
Pressure will increase if temperature increases, assuming that the container remains at a constant volume. Because the molecules are moving faster, they collide more often with the walls of the container, so the pressure goes up.\n\nThis is demonstrated by Gay-Lussac's Law, which states that that for a fixed amount of gas (fixed number of moles) at a fixed volume, the pressure is proportional to the temperature, so:  P/T = constant.\n\nIn case you need to know this for a test, and easy mnemonic  is "Pete and Tom are Gay" (Pressure and Temperature are proportional according to Gay-Lussac).
482741
You need a few different constants to figure this out.\n\nWater has a Specific Heat of 1 kcal/kg/°C\nThat means it needs 1 kcal per kilogram to heat the water 1 degree celsius\n\nIt has a density of 1000 kg/m3\n\nIt has a Latent heat of Fusion of 80 kcal/kg (melting) and 65 J/kg (freezing)\n\nSo if you know the volume you can work out the weight of the water. Once you have the weight (or you already know it), you can work out the rate of the drop (or increase) of water temperature.\n\nOnce you get to the point of freezing, or melting, simply add on the latent heat, and you can then get the total amount of energy required to change the water from one temperature to the next.\n\nSo for example;\n\n1) How much energy to heat 2 kg water from 18°C to 100°C is =>\n\n2kg * 1 * (100-18) = 164 kcal (The heat up curve is fairly straight)\n------------------------------->\n\n2) 2kg of water from 5°C to -10°C is as follows;\n\n(2 x 1 * (5 - -10)) + (65 * 2) = 160 kcal\n\nCurve drops down fairly constantly, then flats out until the water turns to ice, then continues to drop until you are at your temperature.\n\nI hope that helps a little.\n\nIf you want to figure out the times for all this to happen, you need to either know what heat source you are using (for example the wattage of the kettle element), and also you need to figure out the heat losses from the container (e.g. kettle or pipe, etc.)\n\nThese can all then be converted to an equivalent kcal number.
48257
ummm, no, water doesn't always boil at 100C.  It boils at 100c at one atmosphere, water will boil at room temperature in a vacuum.  But, hot water will boil faster, it requires less energy to heat the water, and the less water you have to boil the faster it will reach the boiling point.
765689
Yes you are right to subtract the water pressure from the total pressure.\n\nThen use p1v1/t1 = p2v2/t2\n\nSTP is p2v2/t2\n\nSTP = 1 atm and 0C or 273K\n\nSolve for V2.\n\nKeep units the same on both sides!
1144024
Water acts as the medium in many processes. 25 deg C is the room temp.
780627
WATER BOIL AT:  27*C at 26.739 (pressure(torr)  thru 30*C at 31.824\n pressure(torr)
432890
Pure =32 degrees F\nSalt = depends on the salt concentration, ocean water is -40 degrees F
734516
Easy.  \n\nYou have q = mass * specific heat * change in temp.\n\nFind the specific heat of water and your set.  Q is joules so you need to convert.
132159
Open up your thermodynamics book and do your own homework.\nHint.. add the energy required to raise the temp from 100C to 112C plus the energy used to change state, plus the energy used to go from 76C to 100C.  You should have all the info you need in your problem, and the energy required to do the change is in your book.
341668
Water evaporates at 100 degrees, celsius, so the liquid water you touched wouldn't be any hotter than that. The metal, on the other hand, does not. So it would be as hot as the object which is heating it.