How would you calculate a heat load to cool or heat a room?
960599
If it is a residential application:  850 sqft/ton\nIf it is a commercial application:  350sqft/ton\n\nEx.  1700 sqft / 850 sqft/ton = 2 tons of air conditioning
945879
The first method would be to compare you utility bills to see if there is a drastic change (or any) between last year at this time and the most recent bill with the insulation on.  \n\nFor an exact calculation, you need the following:\n\nThe R value of the insulation\nLength of pipe insulated\nTemperature of the water inside the pipe\n\nThen plug them into this equation:\n\nQ=UA(Tinside - Toutside)\nwhere Q is the amout of Btus \nU is the thermal insulation (or lack there of) expressed as =1/R\nTinside = water temperature\nToutside = air temperature (average) where the pipe is installed\nA = the surface area of the pipe\n\nOnce you get that for the insulation, do it once with out the insulation.  For the second R value without the insulation, use a R-value of 0.5.  The two different Q's you get will be the amount of Btu's you saved.  \n\nTo convert that to $, you need to convert the Btu's (Q) to kW.  Divide the Btu's by 3.412 to get watts.  That will be the amount watts you saved at your hot water heater/boiler or whatever is heating the water (assuming it is electric).  You then look at your electric bill to determine the cost $/W or $/kWh.  Take the number you got for watts and convert to kilo Watts by dividing by 1000.  Then multiply it by the cost off of you utility bill for $/kWh.  \n\nThis is all for a one hour period.  You need to calculate that for a month, day, or year (720, 24, 8765 hours respectively).  You now have what your savings for this project was.
615198
use the formula q=m*s*(t2-t1) where q is energy in heat,m is mass in grams,s is specific heat in joules/gram celsius,t1 is initial temp,t2 is final temp.\nsubstituting in formula,\n42,000=900*4.186*(t2-35)\nt2 works out to be 46.14 celsius.\nhope this helps.
897593
If you are doing this in terms of the system,\n\nsince the system is losing energy (delivers heat) then the amount of energy is "-1275 J".  The surrounding add heat (perform work on) to the system, so the amount of energy is "+855".  The internal energy of the system is the sum of the energy loss and the work done to the system, so -1275 J + 855 J = -420 J.\n\nIf you are doing this problem in terms of the surroundings, the all the signs are reversed and you should end up with +420 J.
474430
heat from your chamber wall
920164
Does the "promotional litearture" keep you warm?  Try burning it, that might produce some heat.\n\n$300 is too much, you need that money to pay your heating bill.  Anyway, I seriously doubt quartz is any more efficient than a cheap space-heater.  They're both 100% efficient at converting electricity into heat.  To improve on that you need a "heat pump".  \n\nThose are just an air-conditioners that are turned around, so that instead of pumping heat out in the summer, they pump heat inward in the winter.  To use one you need a source of heat that is above 40-50 deg.F, such as well water.\n\nYou can als:\ndress more warmly\nturn up the thermostat\nget another space heater\nadd insulation to your house
867420
about heating maching bring a thick wire and wrap it around a spoon then the tips connect them into electric source (big one) dont do it alone it gets really hot ...ok\nabout the cooling machine i dont know
485249
It is not clear exactly what you wish to achieve but the natural way to ventilate a space is to have inlet openings near the bottom of the space (with in a foot or so of the floor) and outlet openings at the peak of the room. The all of the inlet openings should be evenly space around the perimeter. The outlet should run the full length of the roof peak if possible. The area of the inlet and out openings should be equal with the outlet opening slightly larger in area if possible. \n\nThis arrangement permits convection currents to keep fresh air coming in near the bottom of the space. There will be an area in the center of the floor that will not be very well ventilated unless the space is relatively narrow.
737811
You use the formal \nH= mcdeltat\n\nmanipulate it to be...\n\nc (specific heat capactiy) = H /(mdelta t)\n\nChange temperature to Kelvin,....4degrees = 277Kelvin\n\nC=221/(55X277)\n\n\nC= 221/15235\n\nC=0.0145 (in JOULES)
1017637
I will assume you want to find the sq ft of a house or building. You just multiply the length by the width. If the build has several different dimentions you will have to figure each area seperately and then add up the totals.
554422
Get a couple of fans and make a "wind tunnel" effect, to move air thru your room, and not just push it around.  By pulling air from another room, you should be able to cool down yours.\n\nI found that putting a bowl (metal works best) of ice in front of a fan will cool things down until the ice melts.
936508
Neither. Let's have a given that 1 fridge is empty and at 40 degrees inside. The other fridge is full and at 40 degrees. The room that these 2 fridges are in is at 70 degrees. The difference in temperature is the cause for heat to flow from hot to cold. The fridges have insulation that has a certain resistance to heat flow. The heat will flow through the walls of the fridges the same, so then Q(heat flow)=A(area)x(T1(70 degrees)-T2(40 degrees))/R(resistance). The area of heat loss for both fridges are equal,the resistance of both fridges are equal, the temperature differential for both is equal,  so the amount of electricity needed to keep the fridges at 40 degrees is equal.
353934
I presume you're asking about an air conditioner.\n\nAir conditioners are rated in terms of the equivalent cooling capacity of tons of ice.   So, a one ton air conditioner can cool as much as a ton (2000 lbs) of ice melting in a 24 hour period.   A home air conditioning unit is typically in the range of 1-3 tons of capacity.  One ton is also equivalent to about 12000 BTUs (British Thermal Units) per hour of cooling.
470685
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.
801436
Use specific heat capacities and latent heats of these materials along with your head to get water at 0C
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.
1037873
Depends on what you mean by types.\n\nDo you mean the type of material used in constuction or the type of activity the building is intended for.\n\nMost large building use a chilled water system where they use chillers to cool water down to about 42 deg F then pump the water throught the building to air handler units which blow air across the coils filled with the water.  This heats the water, but cools the air.  As the air is circulated throughout the area the water is returned to the chiller to be recooled and the process starts all over again.\n\nHowever, an important point is capacity.  That will be determined by both what the intended use of the building is and what sort of material is used.