The formation of condensation on a cold glass of water will cause it to warm up faster than it would have otherwise. If 6.6 gm of water condenses on a 109-gm glass containing 285 gm of water at 5.0 deg C, what will be the final temperature ? Ignore the effect of the surroundings. Express your answer to the nearest tenth deg C.
123518
The subsequent evaporation will cause the temp to remain less than .1 deg c diff.
488561
No, you can't say nothing has happened.  Water evaporates until the system reaches equilibrium.  As the temperature rises and falls, the air in the glass will contain a little more or less water, but there is no major change.  If condensation occurs, the water falls back into the glass.  It is a closed system in equilibrium.
469078
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
389673
You asked this question twice, and I posted a detailed solution in answer to your other question.  I know your teacher's answer is 49 degrees, but if you go to the nearest tenth of a degree, 48.9 is a better answer.
505414
The potential energy mgh of 1 gram of water 0.001 x 9.813 x 50 = 0.49 Joule. \nThe specific heat of water is 4.2 J/g/K, therefore 0.49 Joule raises the temperature of 1 gram of water 0.49/4.2 = 0.12 degree.\n\nThe  temperature of the water at the bottom of the falls is 10.12 C
174357
There are multiple steps here\n(1) Ice from -25 to ice at 0\n(2) Ice at 0 to water at 0\n(3)water at o to water at 100 (l)\n(4) water at 100 (liq) to water at 100 (vap\n(5) water  from 100 to 121\n\n(1) H1 = m x S x t\n(2) H2= m x l (latent heat of fusion)\n(3) H3= m x sx t\n(4)H4= m x L (latent heat of vapourization)\n(5) H5 = m x s x t     (all "t " corresponds to change in temp)\n\n(1) H1=70 x 2 x 25=3500\n(2) H2= 70 x 6.01x1000 =420700\n(3)H3= 70x 4.2 x 100 = 29400\n(4)H4 = 70 x 40.7x 1000 = 2849000\n(5) H5= 70 x 2 x 21 = 2940\n\nTotal Heat energy= 3305540 j/mol = 3305 KJ/Mol\n\nThis is the solution... I have a question: You gave the values of Phase change enthalpie in KJ/Mol... If they are correct , then this anser is perfect but if they are in J/mol then the answer u will get around 39KJ/mol\n\nThis solution helped you?
302249
She will need enough heat to raise the temperature of the ice 20 degrees and then to fuse it all at a constant temperature.  Water's heat of fusion would not have a temperature term as it is a constant temperature process.\n\nSo, it would be:\n2100 J/kgC * 2.0 kg * 20 C + 335,000 J/kg * 2 = 754,000 J or 754 kJ
398824
1) Raise the temperature from -20°C to 0°C:\nMass of ice * specific heat of ice * temperature rise\n10.0g * 2.1 J/g K * 20° = 420 J\n2) Melt the ice:\nMass of ice * Heat of Fusion\n10g * 333J/g = 3330 J\n3) Raise the temperature from 0°C to 100°C:\nMass of water * specific heat of water * temperature rise\n10.0g * 4.2 J/g K * 100° = 4200 J\n4) Evaporate the water:\nMass of water * Heat of Vaporization\n10.0g * 2260J/g = 22600 J\n5) Raise the temperature from 100°C to 120°C:\nMass of steam * specific heat of steam * temperature rise\n10.0g * 2.0 J/g K * 20° = 400 J\n\n6) Addition: 420 + 3330 + 4200 + 22600 + 400 = 30950 J
500905
Nothing chemically.  The heat from the liquid water will cause the CO2 (dry ice) to sublimate into vapor.  The mist you see is water vapor condensing in the cold temperatures.  Is that what you wanted to know?
15915
The formula for calculating specific heat is q=mC(T2-T1). Where q=heat exchange, m=mass, C=specific heat, T2=final temp, and T1=initial temp.\n\nSince you have two materials inside your container, your formula would be q=[mC(T2-T1)]for the water + [mC(T2-T1)] for the metal.\n\nThe container is insulated so there is no heat exchange between the outside environment and the materials in the container. So q=0.\n\nThe formula reduces to 0=[mC(T2-T1)]for the water + [mC(T2-T1)] for the metal. \n\nOr [mC(T2-T1)]for the water = -[mC(T2-T1)] for the metal\n\nC of metal= [mC(T2-T1)]for the water/[-m(T2-T1)]for the metal.\n\nC of metal= [(185g)(4.184 J/gK)(303.75K-298.15K)]/\n            [-(36g)(303.75K-451.15K)]\n\nC of metal = 7775.313J=7.775kJ
431293
Conservation of energy.\n\nCalculate the amount of heat stored in the 28ºC water using 0ºC as the starting point.  This energy must first melt the ice, for which you'll need to use the heat of fusion of water.  The remainder of the energy will warm up the entire amount of water.
342950
the first thing you do is figure out how much heat it will take to melt the ice, this is done by multiplying the quantity of ice by the heat of fusion\n\n10g x 335 J/g = 3350J\n\nnow figure out how much the water cools off to give the ice that 3350J that it needs to melt, this is done by multiplying the amount of water by the specific heat and then dividing this into the energy transferred\n\n100g x 4.18J/gC = 418 J/C\n\n3350J / 418 J/C = 8 degrees C\n80-8=72C\n\nso now you have 100g water at 72C and 10g water at 0C, so you can figure out what this mixture will equilibrate to\n\n100 x 72 =7200\n10 x 0 = 0\n7200 / 110 = 65.4C will be the final temp of the mixture\n\nso the answer is D
493759
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?
20651
H = mct + mL\ntherefore heat needed to take 22.7 g ice from -22.19 to 0oC\n\nH1= 28.7 x 22.19 x 2.1\nH1= 1337.4\n\nH2 = mL = 333 x 28.7 = 9557.1\n\nH3 = m(water)c(water)t = 28.7 x 4.2 x 100 = 12054 \n\nH4 = mL = 2260 x 28.7 = 64862\n\nH5 = mct = 28.7 x 2.0 x 126.2  = 7243.88\n\nHtotal = H1 + H2 + H3 + H4 + H5\n\nTherefore\n Htotal= 95054.38
574146
Hot water will drop in temperature at a faster rate, but the cold water will freeze first except under certain circumstances in a phenomenon known as the Mpemba Effect\n\nhttp://math.ucr.edu/home/baez/physics/General/hot_water.html\nfor more details on the phenomenon
22461
The same thing would happen if the table were metal!\nHeat energy in the table is absorbed by the cold glass. When the dew-point is reached underneath the table, the humidity in the atmosphere condenses on that surface as well as the glass on top.\n-it has nothing to do with similar materials-
663664
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