using fermi dirac statistics obtain the expression for eletronic heat capacity of metal.this is m.sc question?
147206
were do you start from? ideal quantum gas?\n\nwhat i know is: if you have managed to compute entropy S, heat capacity c is simply:\n\nc = T ∂S/∂T
729915
E = cp*m * ∆T; you have cp (specific heat), m and ∆T.  Find how many moles 1kg of iron is, and subsitute that for the mass in the formula for cp.  You can get that information here http://en.wikipedia.org/wiki/Iron
432258
Since it is at thermal equilibrium, the energy (q) of the water must equal the energy (q) of the metal.\n\nTherefore: mcAT = mcAT\n\n(90)(4.18)(0.9) = (11)(87.2)(c)\n\nSolving for c, the specific heat of the metal would be 0.353 J/g*oC
437730
See http://members.aol.com/logan20/elecspon.html for the equation to use.
309325
Since m*C*dT=Q;\n\nHeat capacity=135000 J/(5100 g*(31.5K - 18K))= 1.96 J/gK\n\n(K for Kelvin)
540767
The conductivity of aluminium is required. A detailed solution of the Heat equation (A pde) is necessary to get an accurate answer. In the absence of any simplifying assumptions, this problem is not trivial
158584
Try,\n\nhttp://www.personal.psu.edu/faculty/d/x/dxh40/MBLemma.pdf
954813
Yeah, it's E=mc^2 and it relates energy and mass.  E is energy, m is mass, c is the speed of light (celeritas).  The equation could come useful in physics, like atomic physics.\n\nWhat did you wanted to know?
52130
E=mc∆T is the equation you want to use for this situation, where E is energy, m is mass, c is the specific heat, and ∆T is the change in temperature.  Now, the heat lost by the alloy is equal to the heat gained by the water, right?  So look what we can do:\nmc∆T=mc∆T\nwhere the left side of the equation is talking about the metal, and the right side of the equation is talking about the water.  Let's start pluggin in values.  We'll leave the left c value as a variable because that's what we're solving for.  \n36.0 g *c*(178 oC -30.6 oC)=185.0 g*(4.184 J/g*K)*(30.6 oC-25.0 oC)\nThe trickiest thing about that step is the temperatures: the metal started at 178 and goes down to 30.6; the water starts at 25 and goes up to 30.6.  \nNow evaluate that equation and solve for c. \nc=.817 J/g*K
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
1045656
An electron passing through a potential difference of 150 kV picks up 150keV of energy.  So:\n150keV=h*c/lambda\nlambda =h*c/150keV.\nh=6.626*10^-34 J*sec\nc=3*10^8m/sec\n1eV=1.6*10^-19 J\n\nSo, lambda=((6.626*10^-34)(3*10^8))/((150000)(1.6*10^-19))\n\nlambda=8.28*10^-12 m, or 8.28 nm
794988
Energy radiated per second is given by\nE = e*sigma*A*T^4   where \ne = emissivity; \nsigma = Stefan's constant = 5.67 * 10^-8 W/m^2/K^4\nA = surface area\nT = temperature in kelvin.\nSubstituting the various values,\nE = 249 * 10^-8 Watts.
963248
The heat given off by the burning reaction (delta E) is equal to the heat gained by the calorimeter.  (m * change in temp)\n\nYou can calculate the heat released by burning.\n.553 g with a molecular wt. of 88g/mole \n(.553 grams) / (88 (grams / mole)) = 0.00628409091 moles\n(975 (kJ / mole)) * 0.00628409091 moles = 6 126.98864 joules\n\nYou can calculate the heat absorbed by the water (3000 g * .356)\n3 000 * .356 = 1 068 joules\n\nyou can calculate the difference, it must have heated up the apparatus.\n6 127 - 1 068 = 5 059 joules \n\n5059 joules = x grams * .356\n5059 joules / .356 = 14210.67 grams of water\n\nThe metal apparatus was had a heat capacity equivalent to 14210.67 grams of water.
14789
A general form of the equation I beleive you are looking for comes from a basic understanding of thermodynamics.  I am assuming that your ct=(Tfinal-Tinitial)...\n\nSo a general form of the equation would be:\nmCp(Tf-Ti)=Q \n\n"Cp is the specific heat"\n\nWATCH UNITS!!!!!  \n(I am ignoring the conversions in this statement)\n55g*Cp(67degrees UNITS)=343 cal \n\nSolving this equation will give you Cp ... just make sure that you watch units.
294164
Thw formular to use here is Q = mcT where Q = quantity of heat, m = mass, c = specific heat capacity, and T is the change in the temperature, thus here Q = 2.0g*(30-25)*0.90 = 9J. Notice that I sed Joules as the unit.\n\n2.For this one, a law which i cant remember the name right now is nedeed. Here we have C(s)+O2(g)=CO2 Enthalphy -390KJ\nand also Mn(s)+O2(g)=MnO2(s) Enthalphy = -520KJ\nFor this law to be appplied, we have to write the second equation the other way to become MnO2= Mn+O2 Enthalphy = 520KJ. Notice that i changed the sign when i reversed the equation. Then O2 could be canceled out on the left side of the first equation and on the right side of the second equation. This was we are adding. Finally adding, we get MnO2(s)+C(s) = Mn + CO2 Enthalphy = -390KJ+520KJ = 130KJ.\nand thats the correct answer
984240
The atomic radii for Cr, Mn, Fe, Co, Ni, and Cu are\n\nCr: 125 pm\nMn: 137 pm\nFe: 124 pm\nCo: 125 pm\nNi: 125 pm\nCu: 128 pm\n\nSo, you can see, there is a bit difference between their sizes. Although among them, Cr, Co, Ni, and Cu do have very similar atomic radius. \n\nThe reason that these elements have similar radius is due to their electron configurations. So, let's look at their corresponding electron configuration first (hope you know what I am referring to) :\n\nCr: [Ar] 3d5 4s1\nMn: [Ar] 3d5 4s2\nFe: [Ar] 3d6 4s2\nCo: [Ar] 3d7 4s2\nNi: [Ar] 3d8 4s2\nCu: [Ar] 3d10 4s1\n\nAs you may realize, the major difference between their electron configuration is mainly in the "3d" orbital, which has a quantum number (n) of 3; this 3d orbital has a smaller radius than the next orbital, 4s. The reason why 4s orbital is filled BEFORE 3d is due to the lower energy level of 4s compared to 3d.\n\nIn short, the major reason why these elements have similar radii is because the newly added electrons are not added to the "outermost electron shell", instead, they are added to the "inner electron shell". Therefore, no SIGNIFICANT changes to their radius.
414002
one cannot measure electronegativity.it has been given on the Pauling scale given by Pauling .and charge can be determined by the valency of the element.