.When a heat of Q is supplied to a diatomic gas in a certain process it does a work of Q/4 . Molar heat capaci
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'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
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Your data tells you how much heat is given out when 2 x 46g of NO2 is formed. Now scale up to the quantity given.
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HEAT ENERGY=mQ(t2-t1)\n\nM-MASS OF SUBSTANCE\nQ-SPECIFIC HEAT CAPACITY\n(t2-t1)-CHANGE IN TEMP.
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I think it is this:\n\n            \nC12H22O11(s) --- HNO3---->  12C(s) = 11H2O(g) + heat
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Go here to find out about  "MOLAR ENTHALPY"\n\nhttp://www.carolina.com/chemistry/activities/Determining_Molar_Enthalpy_of_Solution.pdf
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The heat energy Q  that must be absorbed by a mass m with specific heat c raises the temperature of the mass from temperature T1 to temperature T2: \n\nQ = mc(T2-T1)\n\nThis assumes there are no other energy losses or gains to the system.
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um.. this is a physics question, but i'll answer it~\n\nQ=cm(delta)t, therefore, Q=(4.184 J/g C)(75g)(60-35 C)\nQ= 10983 Joules
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heat is a form of energy.
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Water acts as the medium in many processes. 25 deg C is the room temp.
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Your question is not clear but I take it you want 2 know what happens when heat is applied to a chemical reaction. \n\nHeat increases kinetic energy in molecules. If you apply heat, bombardment of molecules increase and therefore the reaction is favoured. In terms of equilibrium in reversible reactions, the forward reaction is favoured.
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C6H12O6(s) ---H2SO4--> 6C(s) + 6H2O(in H2SO4 solu.) + Heat
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Quantity of energy lost by the water:\n\nQ = (m)(deltaT)(c)\n    = (50.0g)(2.00 C)(4.18 JK-1g-1)\n    = 418 J\n\nSince the water got colder, the reaction was endothermic, and the enthalpy change for the reaction is +418 J.
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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.
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Quantity of heat (Q), measurement = joule (J)\nSpecific heat capacity in liquid, measurement = c/(kJ kg^-1)
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Specific heat does not apply to reactions, only actual chemicals.  You are probably thinking of the heat of reaction, or enthalpy.  The change in enthalpy of a reaction (written as delta H, delta as the Greek letter that looks like a little triangle) flips its sign when the reaction is run in the reverse direction.  So a positive delta H value becomes negative when the reaction is run in reverse direction, and a negative delta H value becomes positive.
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From the first law of thermodynamics:\n\nQ = ΔE + w, so ΔE = Q - w\n\nwhere ΔE the change in internal energy, Q the amount of heat and w the work.\n\nNow, the change is isobaric, so: Q = nxCpxΔΤ and Cp = 5R/2 for an ideal gas.\n\nQ = 5nxRxΔT/2 = 5pxΔV/2  (because nxRxΔT = pxΔV)\n\nw = pxΔV\n\nΔE = 5pxΔV/2 - pxΔV = 3pxΔV/2\n\nΔE = 3x1x(0.980 - 1.350)/2 = -0.555 L.atm\n\n1 L.atm = 101.4 J (approx) so\n\nΔE = -0.555 x 101.4 J = -56.28 J (approx)
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Friction, Heat, Acid.