how much energy is required to bring 750g of water from room temperature (18 dgreesC) to boil point?
454337
750g = 0.75 kg\nDelta T = 100-18 = 82°C (or K)\n\nEnergy required = 0.75 x 82 x 4200 = 258300 Joules\n\n258.3 kJ\n(also 0.07175 kWhr, or 244.8 BTU)
705505
The specific heat of water is 1 calorie/gram °C = 4.186 joule/gram-°C.  If you want to use 4.18, that is okay.\n\nThe heat added = the specific heat of water (c) times the mass of the water (m) times the change in temperature.\n\nQ = cmT or\n\n= 4.18 joule/g-°C * 125 g * 2.44 °C\n= 4.18 * 125 * 2.44 joules (you can do the math)\n\nThis is for 1.75 g of CaCl2 which has a molecular mass of 111.0 g/mol, so the number of moles is:\n\n1.75 g / 111.0 g/mol = 0.01577 moles of CaCl2.\n\nDivide the heat change (in joules) above by 0.01577 moles to get the heat per mole.
123942
The normal boiling point is, of course 100 degrees Celcius at standard pressure at sea level (760 torr). \n\nHowever, boiling point depends on several factors:\n\n1. What is the pressure? At higher altitudes, water boils at lower temperatures.\n\n2. Is it pure water, or water containing dissolved materials and other impurities? For example, dissolved salt can raise the boiling temperature by a few degrees.  \n\n3. Water can be superheated, in a microwave, and be above its normal boiling point without boiling. People have been scalded by this.
693456
That would be relative to the amount of BTU's introduced in the heating process. A BTU is the amount of energy (heat) required to raise the temperature of one pound of water by one degree fahrenheit. Pressure is a factor as well... water boils at 210F in Denver, Co.
870267
.4084c(80)+.4084(heat of vaporization)\nc is specific heat of water\nthat will give you J/KG C   si units
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?
872765
It won't matter how much you started with or how much higher the solution was heated.  When the solution cools to 20C it will hold 0.22g/100mL of solvent (water).\n\nSo if you start with 1.75g and 0.22g remain in solution, the amount crystallized at 20C will be the difference, 1.53g.
778647
By fractional distillation.\nDEG distills at 93 'C, you can collect first and then collect water at 100 'C.
646478
Also watts, joules, and occasionally Calories (amount of heat needed to raise one liter of water one degree centigrade.)
712695
At 100 degrees centigrade, water boils.\n\nAs for 70 degrees centigrade, that equals 158 degrees Farenheit.
328951
The latent heat of fusion for water is 335 Joules per gram. so thats exactly how many joules of energy you need to melt one gram of ice
141221
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.
939108
Given:\nPV = nRT, 862 torr x (1 bar/750torr) = 1.15 bar, 117 C +273 = 390 K, CH4 = 16 g/mol\n\nn/V = P/(RT)\n\nn/V = (1.15bar)/[(0.08315 L bar/K mol)x(390K)\nn/V = 0.035 mol/L x (16 g/mol) x (1L/1000mL)x(1mL/1cm^3)\ng/cm^3 = 5.67 x 10^-4 g/cm^3
872856
It will eventually melt at 2572C, and it will boil at 2850C.\n\nHeat does not cause it to react, typically.
945411
1 calorie is the amount of energy required to raise a gram of water 1k. 1k=1.8F.\n\nSo if you're increasing from 38 to 98.6F that's an increase of 60.6F= 33 and 2/3 K.\n\nSo total energy needed is 33.66 calories/gram.
49337
At room temperature, 0.968 g/cm³
209192
Boiling Point: \n204C (399F) \nMelting Point: \n100C (212F) \n\nCHECK IT OUT BUT I THINK ITS CORRECT