What are the thermodynamics of making ice cream?
918443
Making ice cream is just a matter of freezing the mixture.  You use a mixture of ice and salt (because it has a lower temperature than just ice) to remove heat from the ice cream mixture. So, the ice cream mixture loses heat until it reaches its freezing/melting point, and then continues to lose heat until it is the same temperature as the ice/salt mixture.  That mixture gains heat from the ice cream mixture, and the ice melts, dissolving the ice.\n\nHope this helps...
62689
specific heat of steam is 2.02 J/g*K\n\nU steam= 215000J - 75.0g*2.02J/g*K(25K)= 211213J\n\nU liquified = 211213 - 40,700J/mol*mol/18g*75g = 41629J\n\nU water @ 0 C = 41629J - (4.18J/g*K)*75g*100K = 10279J\n\nU freezing = 10279J - 6000J/mol * mol/18g * 75g = -14721\n\nso you have some water and some ice, or liquid and solid phases.
174163
thermodynamics?  heat transfer?  physics?
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
169447
Adding salt to ice when making homemade ice cream lowers the freezing point of water.  This means that when you add the salt to the ice, the ice is forced to remain at a lower temperature to stay frozen. That makes the ice colder than 0 degrees C, the normal freezing point of pure water.  \n\nLowering a freezing point is a kind of property called a "colligative property," which means that how low the freezing point goes is not dependent on what kind of particles you're using but how many particles you're using.  The more salt you use, the lower the freezing point of the ice becomes.  \n\nYou have to make the ice colder than 0 degrees because otherwise, the ice cream cannot get cold enough to freeze.  You just get left with a milk-cream mixture.\n\nThe dissolution of table salt (NaCl) is NOT significantly endothermic enough to freeze ice cream.
792555
Modern Refrigeration. \n\nSource(s):\n\nhttp://www.constructionbook.com/xq/asp/q...
784478
Yes,\n\nAccording to the laws of thermodynamics the heat required to melt the ice has to come from somewhere.  In this case the heat from the liquid is transmitted to the ice causing it to melt.  (liquids are warmer than solids).  This is the classic case of entropy (http://en.wikipedia.org/wiki/Entropy) this action represents the conservation of entropy, or the conversation of energy.\n\nEntropy is the dispersal of energy that is lost from a system.  Friction is a classic example; as an object moves it has to overcome the force of friction, thus requiring more energy.  This extra energy is lost as heat, and the item that you are moving something over or through absorbs this heat.  Due to entropy it spreads the heat out through the entire system and from there to the entire universe.\n\nWhen something is warmer than absolute zero then the molecules inside it are vibrating; which is kinetic energy.  The faster the molecules vibrate the warmer the object is.  All substances have a point where the vibration becomes fast enough that they turn into liquids, then gases, then plasma.  As an object cools, the kinetic energy is lost from those vibrating molecules and dispersed throughout the universe.\n\nAccording to Wikipedia:  http://en.wikipedia.org/wiki/Law_of_thermodynamics#First_law\nAccording to the first law of thermodynamics:  “Energy cannot be created or destroyed; rather, the amount of energy lost in a process cannot be greater than the amount of energy gained.”  There is your conservation of energy.\n\nAccording to the second law of thermodynamics:  “There is no process that, operating in a cycle, produces no other effect than the subtraction of a positive amount of heat from a reservoir and the production of an equal amount of work”  So the heat from the liquid is transferred to the ice and it does work by melting that ice (raising it’s temperature).\n\nAccording to the third law of theermodynamics:  “As temperature approaches absolute zero, the entropy of a system approaches a constant.”  Another words the energy spreads out until eventually it is even throughout the entire universe somewhere above the temperature of absolute zero.  Since no system is perfect we are always losing energy from it to entropy; which is then spread throughout the universe.\n\nThere are two theories of what happens next.\n\n--In the Osculating Universe theory the black holes absorb all the energy of the universe along with all the matter then they absorb each other creating a super-super-massive black hole.  This black hole cannot sustain itself so it explodes resulting in the big bang.\n\n--In the Steady State Theory the energy dispersion caused by entropy spreads out across the entire universe and it then dies.  For this theory to work black holes would have to have a limit to the amount that they can absorb and either explode or just disappear.  I don’t know of any existing theory that can explain this limit.  Recent discoveries have shown that at the heart of most galaxies (including ours; the Milky Way Galaxy) there is a super massive black hole.  But, for some reason our super massive black hole and those in a few other galaxies are dormant right now.  Can a black hole reach a point where it is full, and stops eating?  Since it can happen then we might end up with a universe populated with dormant black holes and nothing else.  Everything would have dispersed falling into the ultimate state determined by entropy.\n\nSteven Hawking has flip-flopped on the idea, but then after he thought black holes existed he doubted they did exist until it was proven that they do exist.  With the inclusion of black holes then it seems that the Osculating Universe is the most probably model, but we don’t know about Dark Energy or Dark Matter.\n\nWhen astronomers try to measure the mass of the universe they come up short, in fact most of the universe is missing.  Some of it could be hiding in black holes, we don’t know, but not all of it
754837
ice cream melting- physical change as wwhen u freeze liquid ice cream it turns back to ice cream.\n\nsugar dissolving in water- chemical as once the sugar is dissolved ou have to use chemical meathods to get the sugar out from the water.\n\ncandle burning- chemical as when the candle is burning it orduces heat which cannot be got back if we rebuild the melted candle.\n\nbathroom mirror fogging up during a shower- physical as you can wipe the fogg out and if you want it agun have a hot tub of hot water near the galss.\n\nmilk turning sour- chemical as you cannot use phusical methods to turn sour milk into milk.\n\nhope it really helps!!!!!!!!!
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
699816
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
888571
Second law of thermodynamics:\n\nBecause of the 2nd law, you can put a pig into a machine and get\nsausage, but you can't put sausage into the machine and get the pig back."
189642
Fruit, Ice and plain or flavored yogurt.  (I like vanilla).  Blend in blender.  Ta da!
464223
'c' is the right answer.\nuse I law of thermo...\nU=Q-W\nU=Q-Q/4\nU=3Q/4\nBut U=n*C*T where n is moles,C is molar heat cap.for const volume.\nLet us assume that the heat absorbed during the process as\nQ=n* A*T\nwhere A can be defined as molar heat capacity for the process.\nso , n*C*T=3*n*A*T/4\nA=4C/3\nFor a diatomic gas C=5R/2(C=no.of degrees of freedom*R/2)\nA=4*5R/3*2\nA=10R/3
436735
Ice and water are at equilibrium at 32 F (0 C).  A mixture of water and ice cannot be cooled below this temperature.   Salt depresses the freezing point (which is why it is added to icy or snowy roads in winter), allowing lower temperatures to be achieved.  This makes the viscosity build of the ice cream (caused by churning of the mixture) to occur in a shorter period of time, and keeps the ice crystals that form in the ice cream smaller - resulting in a smoother texture.
1164067
Third Law Of Thermodynamics (I am studying that too)\n\nIt is impossible to reduce the temperature of any system to absolute zero by a finite number of operations.\n\nIf you are asking how much internal energy ice has at -273K, then the answer is zero.\n\nGood luck with this.
807714
It depends entirely on the temperatures of the all the components. Presumably you have liquid coffee, a coffee cup of any given material, the air temperature where you are making the coffee and the milk. So all these factors have an effect on each other as far as temperature is concerned. The second rule of thermodynamics is loosely but easily remembered by "Heat as a ruler flows from hotter to the cooler." So if your coffee is colder than your milk, the coffee gets warmer. If the air is colder than the milk, which is colder than the coffee, then the coffee will cool down super fast. If the mug is hot, it will offset the heat loss to the air and milk and so forth. The possibilities can be endless. but a basic answer is: If the coffee and mug are hotter than the milk and the air, and if the air is colder than the milk, then the coffee cools faster before adding the milk, heat flowing from coffee to the air. Adding milk warmer than the air slows the transfer down, as it has added temperature warmer than the air to the equation. If the milk is colder than the air then the heat exchange is speeded up.
413781
http://en.wikipedia.org/wiki/Ice_cream