Please help me solve this problem... it's thermodynamics and I am confused....?
403136
Solve your homework? Heh... I recommend Wikipedia to help you learn about entropy and ectropy. If they are not helpful to you, I'm sure there are some useful links (see the reference part).
252754
ok ill try to explain cuz the guy before didnt. its all basically the same kinda questions.  using pv=nrt, 2 things stay constant  and 2 things change. and you put everything thatz constant on the right side, and then everything thatz changing on the left side. and you get a proportion relation, something like  p1v1 = p2v2  (for number 1 and 2), or p1/t1 = p2/t2. \nand the units are not important at all. the only units that are important is you have to have temperature in kelvin. \n\n1. so pv=nrt, and nrt stays constant in this. if it doesnt say anything about the number of mols changing or the temperature changing, assume its constant. \nso then p1v1 = p2v2. \n(99 kPa)(300mL) = (188kPa)v2\nv2 = (300mL) (99kPa) / (188kPa)\nv2 = 157.9787mL.\n\n2. same thing. p1v1 = p2v2\n(0.988atm)(1L) = p2 (2L)\np2 = (0.988atm) * (1L)/(2L)\np2 = 0.494atm.\n\n6.  so from pv = nrt. u gotta get all the constants on the right side. in this case, the constants are P and n. so divide both sides by p and t. \npv/pt = nrt / pt\nv/t = nr/p.     so if v changes, then t has to change accordingly to make sure that v/t is equal to nr/p at ALL times.\nso v1/t1 = v2/t2. \n(0.67L) / (89+273K) = (1.12L) / t2  \n(89+273K) * (1.12L) / (0.67L) = t2\nt2 = 605.134K\nt2 = 332 degrees celsius.
196178
It is exothermic when the rxn. Heat is negative\nConsider Thermite\n\n2Al(s) + Fe2(O3)(s) –> Al2(O3)(s) + 2Fe(l) ∆Hºrxn=-847.6kJ
144053
Pay more attention in class! And assuming it is an ideal gas, use PV=nRT. Since there is not change in volume you can basically drop it from the equation and make P=nRT. If you like algebra you can leave it in but it will cancel out eventually. You could solve for 'n' first using initial P and T, and then plug and chug.
977712
Considering water and aluminum as one system,\nthe energy balance gives delta U = 0;\n\nTherefore\n\nMwater *Cvwater* (T2 - T1,w) + Maluminim* Cvaluminum* (T2 - T1,a) = 0, where T2  = 50oC, T1,w = 120oC, T1,a = 20oC, Mi = Mass of component i, and Cvi = heat capacity of component i at constant volume.\n\nWith the given information, and looking for the heat capacities in a table, you can solve for Mwater.
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.
1118780
Your question is very open.\n\nheat equations very general can be:\napplied heat = specific heat x mass x increase of temperature\napplied heat = heat capacity x increase of temperature\napplied heat = mass x specific melt heat\napplied heat = mass x specific evaporation heat (at boiling point)\nHeat from a supply = power x time.\nCombinations and others are possible.\n\nwave equation can be x = A sin( 2pi t/T)\nThere are more possibilities.
949133
OMG!\ni have the same question!\nmine is:\nA piece of an unknown metal has a mass of 20.0 g and at 100.0 C is placed in a styrofoam cup calorimeter containing 50.7 g of water at 22 C. If the final temp is 27.2 C, calculate the specific heat capacity of the metal. Assume that all the energy lost by the metal is gained by the water.\n\nok, haha fine, not the SAME question, but close enough. But, i'm uber confused. ><
641078
I think this is how you would solve this problem. Use the equation PV=nRT, re-arrange to solve for V. \n\n(STP = standard temperature and pressure = 1 atm of pressure and 273 Kelvins)\n\nV = nRT/P = (1 mol)(0.0821 L*atm/k*mol)(273)/1 atm = 22.4 L
149603
The amount of heat transfered is\n\n1) directlly proportional to the area of the sheet of given material. (A)\n\n2)directly proportional to the temperature difference on both sides of the sheet. (T2-T1)\n\n3)dirctly proportional to the time.(t)\n\n4)inversely proportional to the thickness of the sheet. (d)\n               \n---------------Q= A * (T2-T1) * t  /  d------------------------
909507
The change in energy is change in heat + change in work (dE = q + w)\n\nIn this situation, the system is releasing heat to the surroundings, , so it is exothermic and q < 0.  q = -350 J.\n\nIn this situation, the system is doing work on the surroundings, w < 0.  To calculate work, W = PdV.  In PV work, since J = kg-m^2/s^2 = Pa-m^3, you need to make sure you convert pressure units from atm to Pascals (1 atm = 101,325 Pa), and volume units from liters to m^3 (1000 L = 1m^3).  \nw = -(1.00 atm)(20.36 L - 1.20 L)(101,325 Pa/1 atm)(1 m^3/1000 L) = -1941 J\n\nThe total energy change is dE = q + w = (-350) + (-1941) = -2291 J
64647
2x (G2 + M2 ---> 2GM)      DH=-280.0*2= -560.0 kJ\n2x (L + 2M2 ---> LM4)      DH=-350.0*2= -700.0 kJ\n   (L2G4 ---> 2L + 2 G2)   DH=122.0*(-1)=-122.0 kJ (change sign when flip over)\n\nAdd them together, we'll have:\n2G2 + 2L + 6M2 + L2G4 ---> 4GM + 2LM4 + 2L + 2G2\nthen cancel 2G2 and 2L on both side, we end up with:\nL2G4 + 6M2 ---> 2LM4 + 4GM\nthe enthalpy of this reaction is, therefore the sum of each enthalpies calculated above.\nDH=(-560.0) + (-700.0) + (-122.0)= -1382.0 kJ (exothermic process)\nhope that'll help!
1032149
Q=MC (delta) T....(5000 G)(.377 J/G degree C)(200 Degree C) which should be.....i get 377 J, which isn't your answer....i just had an exam on this last week, so i thought for sure i am right....the only thing i am not sure about for certain is the specific heat of coal, i got it from a website.....need more info if i am able to help, surely they provided the specific heat of coal....also if latent heat of fusion of vaporization has a part, the answer above is going to be different....sorry dude
1049686
He is asking for mechanism....Mechanism is how it works. This is a phase equilibria problem. Please refer to A Level option booklet of phase equilibria.
925329
You must approach this question with an understanding of energy conservation. Change in Energy = mass*heat capacity*change in temperature, where heat capacity = 1.00 for pure water. By evaluating your conditions with the above definition of thermal energy, it will show that you must supply 2 calories of heat to the system.\n\nIf you were to add another substance to the original system. You would once again approach the problem using energy conservation. The energy of a hot liquid (mass1*heat capacity*change in temperature) + the energy of a cold liquid (mass2*heat capacity*change in temperature) = mass(1+2)*heat capacity*change in temperature.\n\n-Christopher
1064629
find the surface area first (thats easy). Then given that the emissivity of th sphere is .851, and that the rate of change will be the change of temp/time. Since there is no time involved directly, we need to have the Surface area x emissivity rate / the temperature difference (in relation to the time, which is given within the emissivity rate). This should give u the net rate, depending on the emissivity/time relationship. You do the working out, otherwise you won't learn it.
836789
All this problems are solved using the equation:\n\nHeat = M * Cp * DeltaT\n\nPlug the numbers, check wich one is OK.  Oh, and watch the units.