this is for my school project, i am building a hotel but i'm not very keen on boilers and stuff like that. the water will be at approximately 20 degrees celsius and i want to heat  it to 55 degrees celsius. can someone help me please?
213341
You must calculate difference of enthalpy between first and second condition.  Enthalpy depends on  temperature and pressure.And it can be find from thermodynamic tables.\n\nI accept 1 bar for water's pressure \n\n1 bar and 20 degree water's enthalpy = 84 kj/kg \n1 bar and 55 degree water's enthalpy = 104,725 kj/kg \n\ndifferent between inlet enthalpy and outlet enthalpy = 104,725-84 = 20 kj / kg (this heat necessary to increase of 1 kg water's degree from 20 C to 55 C ) \n\namount of water = 6000 lt = 6 m3\nspecific density of water which is at 20 C and 1 bar = 0,0010017 m3/kg\nmass of water = 6 / 0,0010017 = 5990 kg \n\ntotal heat requirement = 5990 kg x 20 kj/kg = 119800 kj \n\nXXX if this proses time is 1 hour then \n\npower or capacity of boiler for perfect situation ( perfect situation = no losses and boiler effiency 100 %) = 119800 / 3600 second = 33,2 kw \nif boiler effiency = 85 % then \nreal capacity of boiler for proces time is 1 hour = 33,2 x (100/ 85) = 39,05 kw \n\nXXX if this proses time is a half hour then \n\npower or capacity of boiler for perfect situation ( perfect situation = no losses and boiler effiency 100 %) = 119800 / 1800 second = 66,55 kw \nif boiler effiency = 85 % then \nreal capacity of boiler for proces time is 1 hour = 66,55 x (100/ 85) = 78,3 kw \n\nif you want to calculate capacity of boiler ,  first you must know proses time and pressure.
483049
You might want to consider using a double wall, with air in between the inner and outer wall. If I remember correctly, air is a very poor heat conductor. Of course, like someone suggested, vacuum would be better, but that's a bit of a stretch. Air would probably work almost as good. The walls should be made out of thermo-insulants (ebonite, I believe)  and/or high heat capacity materials. Covering the outside of the box in aluminium foil would help, if the heat was generated by a source of light (like the sun), but would useless if the heat came from somehwere else. Likewise, painting the box in black would help the water inside stay warm. Covering the inside of the box  with aluminium is as useless as having knees on a fish. Unless you decide you want a lightbulb in the box, for whatever purpose, I'd say it's a waste of time and materials.
55868
The original water temp is 25C, as was given in the problem... I'm assuming you meant to ask what was the initial temp of the water that was added, which isn't too hard. Here goes.\nQ(heat)= m(mass) x c(specific heat) x delta T(change in temp)\nso Q = (150g)(4.something J/gC-look it up)(25C-14C)\nBecause of conservation of energy and assuming ideal insulation, the heat LOST by the 150g of water during the mixing is equal to the heat GAINED by the 200g of water. So, once you have Q, rearrange the equation to look like [delta T] = Q/mc = Q/(200g)(c)\nThat will give you the temperature difference between 14C and the 200g of water's initial temp. Delta T should come out as being positive, since the temp is going up. \nWhen you have more complex mixtures of things with different specific heats and add processes and changes, the math becomes a lot more of a headache but the same basic principles apply just the same.
97931
This depends on what you wanted in the colorimeter in the first place, are you looking at temperature change rates going up, or down?  Either way, if you mix the two, it will skew the results of your experiment.  Hot water, even at 100 C still may have a lot of ability to absorb latent heat of vaporization before it becomes steam.  Your temperature curve will slow significantly or even flatline for a while if you are increasing temperature.  If you are decreasing temperature, hot water has already lost most of the latent heat of vaporization, it has already become liquid, this will impact your results.
501511
A fired heater will probably be the cheapest way.  You could try electric heaters (you can buy the elements from water heaters at appliance repair shops - they use 240V), but the electricity cost will probably be more than for the fuel (if you were thinking gas/petrol, maybe propane or natural gas will be cheaper).  I also don't know if I'd be comfortable putting electric elements in 3500 gallons of biodiesel - I've seen a few kitchen fires, and burning oil is never pleasant.\n\nYou could also consider solar heating - it would be cheaper in the long run, but any efficient installation would have to involve heating panels, piping, pumps, and a lot of water.  You could try painting your tank black, but that probably won't work very well unless you live in the desert or high plains.
614395
Not being a sauna designer, I can only guess that perhaps the heater warms the rocks to a predetermined temperature or for a predetermined time, and the controller that shuts off the heater turns on the water flow for steam production at the same time, the idea being not to have heating and water flow simultaneously for fear of an unhealthy amount of steam.\nFrom the ref. (if one wants "steam production at traditional sauna bathing"): "The electronic control logic’s will not begin the steam production until the selected\nbathing temperature is reached."
476691
Use the fire to heat water in a container to boiling. Pipe it to a steam driven compressor to compress refrigerent to cool the ice chest. Don't forget the condenser, evaporator and expansion valve.
527315
The amount of heat generated will be a function of the resistance of the load on the source, and the surdface area of the plates.  If you have a connected load of high resistance, not much heat will be generated.  If you are short circuiting a 230 volt source capable of supplying high current, I predict you could generate enough heat to produce steam.
667465
Presuming (delta)H is for one mole of CaCl2 dissolving, 13.2 g is 0.119 moles (MW = 111), so the heat given off would be 9.70 kJ.  Assuming the water absorbs that heat, heat = specific heat* (delta)T * mass\n\n(delta)T = heat/(specific heat * mass) = 9700 J/(4.18 * 132) = 17.6 degrees.\n\nFinal temperature should be 42.6 degrees Centigrade.
431523
I think you should break things down and think of the question like this.\n\nThe heat for a process is 250.3J, Its doubled with the same temperrature difference, the temperature difference is 5C.
264363
Okay, I'll try this one -\n\nRun this experiment twice (or more times), each time with a piece of metal having different density.  The metal piece should be of adequate size (a sewing needle isn't going to do it), it needs to be big enough to significantly affect the temperature of the water.  You will need to know the mass of the metal (i.e., weigh it) and have a thermometer capable of reading from room temperature up through boiling or so.  Please note this is not going to give you a highly accurate answer, but can give you a ballpark idea (I'm guessing you're probably not doing this in a 500 million dollar lab).\n\nThe procedure:\n\n1)  Measure and record room temperature as T1.\n\n2)  Weigh the metal, call it M1.\n\n3)  Bring a known quantity of water (weigh it, or measure it and convert to weight, call it M2) to a temperature just short of boiling (say 180F to 200F, use the same temperature for each run of this experiment).  Record the temperature as T2, and remove the heat source (take it off the burner, etc.).\n\n4)  Drop in a metal piece at room temperature.  Allow a couple of minutes for heat to exchange.  Make sure you use \nthe same amount of time for each run of this experiment.\n\n5)  Measure and record the new temperature of the water as T3.\n\n6)  Repeat with your other metal samples.\n\nYou can now estimate the specific heat capacity for each metal.  It will not be super-accurate (heat escapes your system, and temperatures won't have completely equalized), but should give you some idea.  Just assume all temperatures did equalize and that no heat escaped for calculational purposes.\n\nSince (you are assuming) the heat which left the water went into the metal, then the product of mass, specific heat capacity, and temperature change should be the same for both the metal and the water (but in opposite directions).\n\nThis means that:\n\n(M1)(c1)(T3-T1) = (M2)(cw)(T2-T3)\n\nwhere c1 is the specific heat capacity of the metal, and cw is the specific heat capacity of water (look it up on Google to get the units you need).\n\nSolving this, you get:\n\n......(M2)(cw)(T2-T3)\nc1 = -----------------\n........(M1)(T3-T1)\n\n(sorry about the dots, I'm trying to keep everything aligned)\n\nRepeat the calculations for each of the metals.\n\nYou will need to calculate (or look up) the densities of each of the metals.  Then, graph density vs. the specific heat capacities you calculated, and see if your hypothesis is true.\n\nHave fun!
42559
The first thing to check is the temp. setting. No one should have it set higher than 120 degrees. If you find that it is higher than 150, lower it. The relief valve is leaking because the internal pressure is too high,or the heater itself is too old. One thing is for sure, it won't stop, and it will only get worse unless you do something.
545630
It is now clear who is asking for the limit of 3 degrees. Is it the turbine manufacturer, surface condenser manufacturer, or someone else?\n\nI assume the condensate is being pumped. Is it the condensate manufacturer that is asking for the limit?\n\nI would think from a heat conservation standpoint, you would want to keep it as close to the condensing temperature as possible and still be able to pump the condensate.
659625
82-84 degrees is the ideal temp. If you are active (such as swimming laps) 80 degrees is fine. For a pool that is for everyday family fun it should be at least 82.
396048
You just need a few basics to estimate things.\n\n1.  6000 watts is 6 BTU/sec.\n2. 1 BTU heats 1 lb of water 1 degree F\n3. 1 gallon of water weighs 8.3 lbs.\n4. A nominal natural convection heat transfer coefficient is 1 BTU/(ft^2 * hr * deg F)\n\nIf you barrel is well insulated, just use the above (except 4) and you will see that 25 gallons will heat at 0.027 deg F per second.  \n\nIf you arent insulated, you need to use the heat transfer to see what portion of the 6000 watts is lost as the temperature rises.  You can readily do this in general form and the result should be an exponentially asymptotic temperature.
650830
Water acts as the medium in many processes. 25 deg C is the room temp.
156570
first we must know turbine's inlet pressure and temperature (depend on design ) , mass flow of steam to turbine (for 2,5 MW)  and effiency of turbine (for calculate of exhaust conditions )\n\nexample : 100 MW double exhaust steam turbine need to 160 ton steam at 500 C and 50 bar for high pressure   and 40 ton steam at 200 C and 5 bar for low pressure. \n\nthere are many kind of boiler . if you dont know some require value then you can not calculate boiler pressure , temperarture and  stages (eco , eva , superheater) .\n\ni can say only about capacity of boiler  ; \n\ni f we accept turbine effiency = 70 % \nthen amount of approximately steam power = 2,5 . 100 / 70 = 3,571 MW \nand \nİf we accept boiler effiency = 85 % then real capacity of   \nboiler = 4,2 MW \nthis value near capacity of boiler. \n\nboiler inlet condition = 1 bar 50 C (i accept) \nenthalpy of inlet water = (from compressed water tables) = 200 kj/kg \nmass of flow = 5 ton/hour (i accept) = 5000 kg/hour = 5000/3600 kg/sn= 1,39 kg/sc.\n\n3,571 MW = 3571 kw = 1,39 . (hç-200) \n\nhç= outlet enthalpy of boiler \n\nhç = 2769 kj/kg (this value must stay at superheated region) \nwe must look to superheated region at thermodynamic tables. \nand we can find best outlet conditions for 2769 Kj/kg. and temperature can not be high than 500 C for 2769 kj/kg. \nwe must take care to pipe line length because some losses cause to change superheated steam to saturated steam or near conditions. \nif superheated steam early change conditions then blades of turbine can be damaged. at the end of turbine must be saturated steam or little bit big superheated steam for protection of blades. \nfinally \n4 bar , 160 C conditions enough for 2769 kj/kg.\n\n\nsummary : \nboiler inlet 1 bar 50 C \nboile r outlet 4 bar , 160 C \nmass of flow = 5 ton / hour \n\nalso next step , calculate of boiler's economizer , evaporatör , superheater , main steam line pipe dia. , pumps (circulating and feed water) , drums , recirculation lines , condenser , auxilary cooling water system (for cooling of condenser) , gland steam line , regulating system , control syst. , vacuum sys. (if necessary for condenser) , water treatment sys. , chemical dosing points ,  etc. etc. etc.\n\nturbine inlet conditions = 4 bar , 160 C 2769 kj/kg\n\nnet power = 2,5 mw = 2500 kw \n\n2500 = 1,39.(2769-hs) \nhs = 970 kj/kg (enthalpy of at the end of turbine ) \n\nthis entalphy shown turbine outlet conditions \n\n0,5 bar 85,4 C or 0,2 bar 85 C (vacuum)\n\nwater must be cooled up to 50 C (inlet of boiler)