What is the temperature in the empty box if there is one Joules inside?
826041
The answer to your question would depend on (1) whether "empty" means a "perfect vacuum" or filled with air? (2) how is this 1 joule of energy initially distributed inside this box? (3) how large is this box? (4) how much time is allowed to elapse before the termperature is determined? (5) can energy leak out or into the box? (6) is the box under the influence of a gravitational field?\n\nThe temperature of an object is proportional to the average kinetic energy of the molecules in it. In a pure vacuum, there are no molecules. There is nothing to measure the kinetic energy of, and temperature is undefined. If a thermometer were placed in a vacuum, the reading would be a measurement of the internal temperature of the thermometer, not of the vacuum which surrounds it. So having 1 joule of energy inside a box containing pure vacuum would have NO temperature. And if it contained air, then we would need to know the initial temperature, pressure of the air inside the box and also the volume of the box. Then we can approximate the air as an ideal gas and use the following equation to compute the temperature:\n\nPV=nRT\n\nwhere T is the temperature, n is the number of moles of gas and R is the gas constant. Since we know the initial temperture and pressure, and the volume of the box. Then\n\nP(i)V(i)/T(i) = P(f)V(f)/T(f)\n\nT(f) = T(i) [P(f)/P(i)]\n\nSince pressure is dfined as P = F/A (force per unit area) and\n\nE = Fd (force times distance), then\n\nP = E/Ad, where Ad = V volume of the box (here we are assuming that the 1 joule of energy is uniformly distributed inside the box and the volume of the box does not change)\n\nHence, T(f) = T(i) [E/P(i)V(i)]
835018
about 3000 kilo J\nit depends on the atom
512231
20 C is about room temp... it's a gas, clear and odorless.
613974
You need to use the combined gas law:\nP1V1/T1 = P2V2/T2, but your temps need to be in Kelvins, not Celsius.\n\nIt also seems that the volume is being held constant, so we can cancel volume out of the problem.\n\nSo, convertine the temperature first, we get:\n10C + 273 = 383K for our temp1\n\nSo, subbing into the formula we get:\n100mmHg/283K = 900mmHg/x\n\nSolve for x and get your answer.  The temperature will be in Kelvins.
316045
kg m^2\n------- = Joule\ns^2\n\nb = kg m/s
640958
1degree Fahrenheit of heat +1degree Fahrenheit of heat =1degree Fahrenheit of heat bkz there is no heat transfer
602468
Absolute Zero or around 20 K.\n\nMelting point 14.01 K  (−259.14 °C, −434.45 °F) \nBoiling point 20.28 K  (−252.87 °C, −423.17 °F)
41651
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
501191
There is a formula...but this is much easier\n\nhttp://www.onlineconversion.com/temperature.htm
993443
29.5kg x (0.01 m)^2 / (3600 s)^2 = 2.28 x 10^-10 joule = \n0.228 nanojoule (nJ)
214671
work = (200N)*(10.0m)*cos(0) + (50.0N)*(35m.0m)*cos(0)\nwork = 2000 Joule + 1750 Joule\nwork = 3750 Joule
274160
32 degrees F\nor \n0 degrees C
85318
Heat is equal to the change in enthalpy, dH, which is equal to the enthalpy of the products minus the enthalpy of the reactants.\n\ndH=H(products)-H(reactants)\n\n2B + 3F(2) --> 2BF(3)\n\nGet the heat of formation values for each specimen from a chemical thermodynamics table. I don't have a table in front of me so I'm just going to make up values so you'll know how to solve the problem. Let H=100 for B, H=150 for F(2), and H=400 for BF(3).\n\ndH= 2(400)-[2(100)+3(150)]\ndH=800-650=150 kj/mol\n\nI do know that elements alone have a heat of formation of 0, but I don't remember if the element has to be in a certain state or not.
958888
Use the formula:\nQ (heat) = Cp (Heat Capacity) * delta T (Change in Temperature)
827623
F*m = μ*Joules→ m = .25*40000000/1000 = 10000m = 10 km
57809
The SI (International System of Units) unit to express energy, work, or quantity of heat is the Joule (J). 1 J = 1 Newton x 1 Meter.
795841
1 calorie = 4.184 Joules\n1 Joule = 1 watt-second, so\n1calorie (C) = 4.184 w-s\n(100 C/hr)(1 hr/3600 s)(4.184 w-s/C) =\n0.11622... watts