how to calculate the cooling load ?
831873
Is it referring to the heat load calulation for air con
1097124
Aplly the following formula:\n\nG = w * A, where:\n\nG - flow rate [m^3/s]\nw - speed of fluid [m/s]\nA - transversal section area of the pipe [m^2]
963007
multiply the mass of the water by the specific heat and the (temperature required minus the ambient temperature).that will give you the energy in joules.if you know in how much time you want this temperature to be achieved you can calculate the power as power is nothing but the rate at which the energy is dispensed
78350
When the metal enters the water, it begins losing heat to the water. At the same time the temperature of the water rises. This process continues until the temperature of the metal and the temperature of the water are equal st which point (30.6oC) no net flow of heat occurs. The heat lost by a gained system is given by (mass)(Specific heat)(change in temp.)First calculate the heat gained by the water, then calculate the specific heat of the metal.\n\nTemp rise in the water - \n   (change in temp) 30.6 - 25.0 = 4.4\n\nHeat gained by water - \n              \n   (185.0)    (4.184J)/(g oC)  (4.4) = 3.4*10 pwr3 J\n\nThe metal dropped into the water must have the same temp as the water (30.6)\n\nTemp drop by the metal - \n   (Change in temp) = 178-30.6 = 147.4\n           \n Specific heat = \n(heat)/(mass)(change in temp)\n\nSpecific heat of the metal =\n(3.4*10 pwr3)\n(36g)(147.4)\n             \n=0.640 J/g oC\n\nhope this helps. good luck
501191
There is a formula...but this is much easier\n\nhttp://www.onlineconversion.com/temperature.htm
437503
Q = m.s.ΔT\nHeat loss by the copper = heat gained by the metal.\nCopper (heat loss) = 200.0g x 0.38J/g.ºC x (195.5ºC-228.4ºC) = -2500.4J\ndivide the heat by the mass and ΔT of the metal to get the specific heat of the metal:\n2500.4J/(65.0g x (195.5ºC-18.3ºC)) = 0.2171 = 0.21 J/g.ºC (correct SF)
457474
check thermaltake and tagan\n\nhttp://www.thermaltake.com/\nhttp://www.tagan.com/
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
790311
To get the most out of your room air conditioner, you'll need to consider the size of the unit that best suits your needs.\nThe cooling capacity of room air conditioners is measured in British thermal units per hour, or Btu/h. To figure out the maximum cooling capacity for a room, you will first need to calculate the size of the room where the air conditioner will be placed.\nTo do this, simply multiply the width of the room by the length of the room.  Use the following information and formulars to help you with calculating your air conditioning requirements:\n\n\nHeat Load\n\nThe amount of heat generated is known as the heat gain or heat load. Heat is measured in either British Thermal Units (BTU) or Kilowatts (KW). 1KW is equivalent to 3412BTUs.  The heat load depends on a number of factors, by taking into account those that apply in your circumstances and adding them together a reasonably accurate measure of the total heat can be calculated. Factors include: \n\nThe floor area of the room\n The size and position of windows, and whether they have blinds or shades \nThe number of room occupants (if any) \nThe heat generated by equipment \nThe heat generated by lighting \nArea of Room\n\nThe amount of cooling required depends on the area of the room. To calculate the area in square metres:\n\n\nRoom Area BTU = Length (m) x Width (m) x 337\n\n\nWindow Size and Position\n\nIf, your room has no windows, you can ignore this part of the calculation. If, however there are windows you need to take the size and orientation into account.\n\n\nSouth Window BTU = South Facing window Length (m) x Width (m) x 870\nNorth Window BTU = North Facing windows Length (m) x Width (m) x 165\nIf there are no blinds on the windows multiply the result(s) by 1.5.\nAdd together all the BTUs for the windows.\n\nWindows BTU = South Window(s) BTU + North Window(s) BTU\n\n\nOccupants\n\nYou will have to take that into account people who normally working in the space. The heat output is around 400 BTU per person.  \n\nTotal Occupant BTU = Number of occupants x 400\n\n\nEquipment\n\nClearly most heat in a room is generated by the equipment. This is trickier to calculate that you might think. The wattage on equipment is the maximum power consumption rating, the actual power consumed may be less. However it is probably safer to overestimate the wattage than underestimate it.\n\n\nEquipment BTU = Total wattage for all equipment x 3.5\n\n\nLighting\n\nTake the total wattage of the lighting and multiply by 4.25.\n\n\nLighting BTU = Total wattage for all lighting x 4.25\n\n\nTotal Cooling Required\n\nAdd all the BTUs together.\n\nTotal Heat Load = Room Area BTU + Windows BTU + Total Occupant BTU + Equipment BTU + Lighting BTU\n\nThis is the amount of cooling required so you need one or more air conditioning units to handle that amount of heat.\n\n--------------------\n\nGeneral thumb rule \n\n100 sqr ft area with roof height of 9 - 10 ft will need 1 Ton AC unit.
750705
212 F, 100 C,     373 K
827524
Easier compared to one with a higher heat capacity:\n\nH/(m*Cp) = (delta T), where m is the mass of the object, H is the heat that must be added (or removed) to raise (or lower) its temperature, delta T is its change in temperature, and Cp is its heat capacity. From the above formula, you can see that larger values of Cp result in small temp. changes, for the same heat amount.
753535
Just use the formula\n\nE = m*Cp*(Delta T)\n\nE=energy\nm=mass\nCp= Specific Heat\nDelta T = change in temperature\n\nCp= 0.51998 J/(C*g)\n\nGood Luck with the test!
52130
E=mc∆T is the equation you want to use for this situation, where E is energy, m is mass, c is the specific heat, and ∆T is the change in temperature.  Now, the heat lost by the alloy is equal to the heat gained by the water, right?  So look what we can do:\nmc∆T=mc∆T\nwhere the left side of the equation is talking about the metal, and the right side of the equation is talking about the water.  Let's start pluggin in values.  We'll leave the left c value as a variable because that's what we're solving for.  \n36.0 g *c*(178 oC -30.6 oC)=185.0 g*(4.184 J/g*K)*(30.6 oC-25.0 oC)\nThe trickiest thing about that step is the temperatures: the metal started at 178 and goes down to 30.6; the water starts at 25 and goes up to 30.6.  \nNow evaluate that equation and solve for c. \nc=.817 J/g*K
363092
Well I'm not going to do the calculation for you but you can.\n\nDo a web search and you'll find dozens of sites to help you if you don't have a fluid flow handbook.
517321
cfm. Cubic feet per minute, a measure of air flow
174163
thermodynamics?  heat transfer?  physics?
565339
For all practical purposes, the answer is zero, but obviously they want something a little more sophisticated.  Refer to Newton's Law of Cooling:\n\nT(t)=T0+|T1-T0|e^(-kt)\n\nFirst solve for k, using the given info:\n\n6=0+|100-0|e^(-k*6)\n6=100e^(-6k)\n.06=e^(-6k)\nln(.06)=-6k\nk=.469\n\nThen, substitute k back into Newton's Law of Cooling equation, and use the new time of 2 hours, making sure to use 120 minutes.  I believe your answer will be choice (b), but check my math!