KETTLE HAS A FAULT SWITCH Energy used=258300J to bring 750g of water from room temp(18 degC)to boiling point.?
336893
You've used 258.3kJ to boil the water.\n\nIf you use 2400 kJ/kg to turn water to steam, and you waste 50g of water then;\n\n=> 2400 / 1000 * 50 = 120 kJ\n\nTotal Energy used = 258.3 + 120 = 378.3 kJ
982154
calories=the energy needed to heat 1 gram of water 1 degree\nso to get the answer  you take 50 x 7 which gives you how many calories it takes
507893
1 calorie heats 1g of water by 1 deg C.\n630cal heats 630g of water 1 deg C\n630cal heats (630/15)g water by 15deg
762941
The change in heat = specific heat * mass * change in temperature (Q=C*g*ΔT).\n\nAssuming that the Aluminum also ends at 28.3 °C, then the change in heat for the Aluminum is\n\n0.0902 J/g°K * 35.7 g * (28.3°-81.9°)\n\nThe above will be negative as heat was lost from the Aluminum.  The change in heat for the water is\n\n4.184 J/g°K * 75.0 g * (28.3°-24.9°)\n\nThe above will be positive as the water gained heat.\n\nThe Heat Capacity of the calorimeter is the difference between the heat lost by the Aluminum and that gained by the water divided by the change in temperature (which is the same as for the water (28.3°-24.9°).
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.
801823
Quantity of energy lost by the water:\n\nQ = (m)(deltaT)(c)\n    = (50.0g)(2.00 C)(4.18 JK-1g-1)\n    = 418 J\n\nSince the water got colder, the reaction was endothermic, and the enthalpy change for the reaction is +418 J.
741934
This question is not very difficult, but you have to be careful with the unit.\n\nFirst, to solve this question, you have to make the assumption that there is no energy lost to the surrounding. So, this reduces to a two system question: one system releases energy while the other one absorbs energy. In this particular case, the energy released by one system equals to the energy absorbs by another system.\n\nTo picture this, consider the following, one system goes through the following steps:\n\n[100 degC H2O (g)] --- (step 1) ---> [100 degC H2O (l)] --- (step 2) ----> [79 deg C H2O (l)]\n\nThe other system goes through:\n\n[260g, 19 degC H2O (l)] --- (step 3) ----> [260g, 79 degC H2O (l)]\n\nIt is obvious that both the energy involved in step 1 and 2 is negative (release energy) while the energy involved in step 3 is positive (absorb energy). We will now solve this question backwards. Starts from finding out the energy required by step 3, then find out the mass of steam that is needed to do the job.\n\nEnergy absorbed during step 3:\nE3 = m * c * (delta T) = 260 * 4.18 * (79-19) = 65208 J\n\nEnergy released in step 1:\n(you have to first convert heat of vaporization to unit of J/g: 40.7*1000/18 = 2261.111 J/g)\nE1 = m' * l = m' * 2261.111 = 2261.111m'\n(since we don't know how much steam is needed, let's just leave it as a variable)\n\nEnergy released in step 2:\n(the temperature of the same mass as above will be lowered to 79 degC)\nE2 = m' * 4.18 * (100-79) = 87.78m'\n\nTotal energy released = (2261.111 + 87.78)m' = 2348.89m'\n\nEnergy released = energy absorbed\n2348.89m' = 65208\nm' = 27.76 g
780627
WATER BOIL AT:  27*C at 26.739 (pressure(torr)  thru 30*C at 31.824\n pressure(torr)
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.
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.
868501
That depends on the specific heat capacity of the oil and the amount of oil in the drum. A rough calculation could solve this assuming the drum is insulated and the amount of water is small. \nEnergy required=mass of oil (kgs)*specific heat capacity of oil (J/kg/degrees)*70 (degrees celcius)\n\nI have used 70 here assuming 20 degrees celcius room temp and assuming you are talking in celcius also. If not, you must convert 90 fahrenheit to celcius and take away room temp.\n\nNow convert 30 minutes to seconds; 30*60 = 900 seconds\nSo you want the energy required to be delivered within 900 seconds, therefore the power you need in watts is the energy required divided by 900.\n\nThis heater power will work in ideal conditions but should be increased (maybe by a factor of 2) to account for assumptions.\n\nI don't know how oil heating works (Don't blame me if you explode!!!) or what "heater capacity" means but I hope the answer in watts will help you.
870952
To answer this Question you need to use the Conservation of Energy Law.   This takes its simplest form in the idea that:\n\nHeat Lost by the Steam    =   Heat Gained by the cold water\n\nYou will need to say:\nLet L be the unknown Quantity and form an Equation:\n\nHeat Lost by the Steam:\nThe steam (one assumes at 100Celsius) condenses to water at 100Celsius.  In doing this it gives up its Latent Heat of Vaporization (mL)   The water it forms is then cooled to the final temperature giving up this Internal Energy to the System.  The Internal Energy(Heat) gained is given by m.s.dT where dT is the change in temperature.\nNote that we have two  Energies to Add Together here.\n\n\nHeat Gained by the cold water:\nThe Internal Energy(Heat) gained is given by m.s.dt where dt is the change in temperature.\nBoth the cold water and the Calorimeter are heated from their starting temperature(15C) to the final temperature(70.6C)\nNote that we have two  Energies to Add Together here.\n\nNow write down the equation:\nHeat Lost by the Steam (steam + hot water)     =   Heat Gained by the cold water (cold water + calorimeter)\n              \nm1.L + m1.s.dT  =  m2.s2.dt + m3.s3.dt\nAnd solve for the Unknown L
299546
water "boils" when the Kinetic energy (energy of motion and heat) cause the molecules of water to expand or separate far enough away that it changes from a liquid to a gas. The greater the force keeping the molecules together, the more energy it needs. On Earth, the force is air pressure. At sea level where the pressure is greatest, water boils at about 100 degrees Celsius. Up in the mountains at 7000 feet it would boil at 80 -90 degrees or even less. In space even the smallest amount of energy would cause it to "boil".\nIt takes one calorie to raise one gram of water one degree Celsius. If you trap the water in a container, you can get the heat of the water much higher than the boiling point. But it could explode if the container is too weak to handle the pressure. So don't try it.
1165181
It's been awhile since i've done this but i'll give it my best\n\nFirst off, the first energy you need to calculate is the energy needed to raise the temperature of ice to 0 degrees C. so...\n\n(12.37 g)(.48 cal/deg/g)(35 deg) = +207.816 cal\n\nWhen the ice changes to water you have a positive heat of fusion. So...\n\n(12.37 g)(80 cal/g) = 989.6 cal\n\nNext you have to raise the water temperature from 0 degrees C to 100 degrees C. So...\n\n(12.37 g)(1.0 cal/deg/g)(100 deg) = 1237 cal\n\nNext you have the positive heat of vaporization.\n\n(12.37 g)(540 cal/g) = 6679.9 cal\n\nNext once it is steam, you must raise the temperature from 100 degrees C to 146 degrees C. So...\n\n(12.37g)(.48 cal/deg/g)(46 deg) = 273.1296 cal.\n\nNow that you have the energy for each step add them together.\n\n207.816 + 989.6 + 1237 + 6679.9 + 273.1296 = 9387.45 cal\n\nI'm pretty confident in my breakdown but it defintely wouldn't hurt to have someone else check it
975732
Substance: Water\nLatent Heat Melting J/g:  335\nMelting Temp°C: 0\nLatent Heat Boiling J/g:  2272\nBoiling Temp °C: 100 \nHeat of vaporization (kJ/mol): 40.65
770429
12\n\n1 Kc / C per litre per degree (litre=1kg)
185660
Water boils and condenses at 100 degrees Centigrade (100*C) at one atmosphere of pressure.  You have not cooled water but have condensed steam (vapor) by cooling the cylinder.  As heat is removed at 100*C, the piston follows the contraction of the steam but it is not the cause of the condensation (heat is lost to the 25*C heat sink).  Therefore, the piston does no work on the steam and neither adds nor subtracts energy to or from the final hot water.  The process is not internally reversible (like compressing and releasing a spring) as you would have to add heat energy back to the cylinder (perhaps at 175*C?) to produce steam and the (frictionless) piston would move to maintain a balance of atmospheric pressure inside and out as the steam expands.  For the piston to do work on the atmosphere, there would have to be an unbalanced force moved a distance.\n\nOverall, the process is reversible by substituting a heat source for a heat sink provided the liquid/vapor temperature remains at 100*C.