the quantity of heat energy in kilojoules required to vaporize 0.493 mol of liquid water to steam at 100 C?
478007
In liquid state, Water has a temperature absorbtion of 4.186 J/gram.\n\nMolecular weight of water is 18 g/mol\n\nhence 0.493 mol of liquid water is 8.874 g\n\nSo according to the previous, you will need 37.15 Joules or 0.03715 Kilojoules
308806
Consider a liquid (water) and its gas (steam).  To convert water to steam, you first have to heat water to 100 C.  At that point additional energy begins to boil the water.  This is the process where water molecules have enough energy to overcome the attractive force between them (called cohesion, in this case, due to the dipolar attraction between waters).\n\nYou are right in that gases are higher energy (could say hotter), but that's because they've taken the energy from whatever heat source was around when they vaporized.\n\nBack to the water example, water on your skin feels cold because it's absorbing heat from your skin and as the water evaporates, that heat is carried away by the vapor phase molecules (which are moving faster than the liquid form).\n\nThe fast-moving particles from your question have a lot of heat stored, that's why steam at 100 C is more likely to burn your skin than water at 100 C.  It has more heat to transfer.\n\nAs for freon, it depends on which freon you're talking about.  Freon 12 (one of the ones phased out because of ozone depletion) melts at -158 C (Merck Index).
960390
Get an electrical heating element, a DC power supply, a thermometer, a voltmeter and an ammeter.\nPass electricity through the heating element, dipped into a measured volume of water (or salt water) in an insulated cup, and measure the current, the voltage and the time taken. This will give you the heat energy supplied. Measure the temperature rise, and work out how much energy would be required to raise 1g of water (or salt water) by 1 degree Celcius. The answer should be close to 4.2!
569208
You need the specific heat of water, which is 4.184 J/g*C.  Assuming a density of water of 1 g/ml (not always the safest assumption, but close enough), then your 500 ml = 500 g.  You're raising the temperature 35 degrees, so the heat you will need will be:\n\n4.184 * 500 * 35 = 73,220 J = 73.22 kJ.\n\nThe combustion of decane provides 6.78 MJ/mol, or 6780 kJ/mol.  In order to give 73.22 kJ, you would need to burn 73.22/6780 = 0.0108 mol of decane.  Given decane's molecular weight of 132 g/mol, you would have to burn 1.425 grams.
21495
the formula is just the heat of fusion for water (in J/kg) times the amount of water you have.\n\nThe heat of fusion in Joules per kg for water is 333548 J/kg, so you just multiply that by the mass in kg to get the joules, so you'd have 333548 J/kg x 40kg = 13341920J which, in scientific notation would be 1.33 x 10^7 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
639379
The boiling point elevation constant of water is required\n\nThe formula is T = Kfm where m is molality and T is rise in temp.
968956
It appears what you really need is the heat of vaporization.  Cp lets you calculate the amount of heat needed to raise or lower the temperature of a certain mass of liquid, but that's insignificant compared to the amount of heat gained or lost with a liquid-vapor phase change.
539607
This is a 2 part question.  You have to figure the amount of heat to raise the temp of the ethanol then find out how much heat to boil.\n\nHeat energy = mass x change in temp x specific heat  \nHeat energy = mass x heat of vaporization\n\nHeat energy = 21 x (76.29-11) x 2.44\nHeat energy = 21 x 38.56\n\nFind those two values and then add them together.
812576
From the reaction you had heat produced\n\nQ=m*Cp*ΔT\n\nfor water Cp= 1 calorie/(gram*Celsius)\n\nYou have a total of 50+50 =100mL and since d=1.00 g/mL you have 100 g\n\nso Q =100*1*(31.43-25.32) = 611 cal\n\nyou mixed mole HCl= C*V= 1*0.05= 0.05 with\nmole NaOH =C*V =1.1*0.05 =0.055 which means that, since the stoichiometry of the reaction is 1:1, HCl is the limiting reagent\n\n0.05 mole HCl give 611 cal\n1 mole HCl gives x\nx=611/0.05= 12220 cal/mole =12.22 Kcal/mole
1112226
Looking at the phase diagram of water,\nat 3000 psi (~20684 kPa, ~2.1 E6 Pa), water can exist in the vapor phase at rougly any temperature at or above 500 Kelvin (rough estimate based on graph).\n\nThere is no one specific, constant, temperature of steam at a given pressure.
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
675016
This is a calorimetry problem! There is an equation for it:       \nq = mc(change in temperature)\n\nq = the heat required \nm= mass\nc= specific heat \n\nAlso, write out the reaction and products.
819353
The amount of heat required to change 1Kg of liquid into gaseous state is known as latent of heat of vapourisation
178523
Although I know I am not getting this $20.  199 kJ of energy.\n\nIn response to a later answer that claimed I was wrong he was right and wrong\n\n He wrote "first find the amount of mol of nitrogen\nmol = 1/14" But infact the amount was 1 kg and that divided by 14g/mol would be 71.43 moles and multiplying that times his enthalpy of vaporization which is 2.7928kJ/mol would give 199 kJ of energy. So I was right, but wrong with my significant figures because the asker said it was 1.0 kg which would gie me 2 sig figs and the answer would be 2.0*10^2 kJ.
412906
To turn liquid water into steam the weak bonds between the molecules have to be broken and the molecules accelerated enough to leave the water surface and become vapour. This requires energy supplied by heating. \n\nNow add salt.\n\nThe salt consists of NaCl crystals. In water these dissolve into Na+ and Cl- ions. These ions will become surrounded by water molecules that are attracted to them, forming large clumps of molecules. To break these bonds requires more energy than in the case where the salt was absent.\n\nWe have to supply this extra energy and the result is boiling at a higher temperature.
677623
False\n\nWhen steam is cooled to water, both can be 100 degrees C, but steam must give off energy to become a liquid.  The opposite is easier to understand.  When you are boiling water, the steam that escapes will be 100 degrees C, but will have much more energy than the 100 degree water.\n\nIn both of these cases, the energy put into or taken out of the water is used to change its phase, without changing its temperature.