What is the temperature in the emoty box, if the energy is 10 joules inside the box?
918910
See if this helps:\nPV = nRT = NkT, where P is pressure, V is volume, n is the number of moles, N is the total number of molecules, T is temperature, k = 1:38 £ 10¡23 J/K is the Boltzmann constant and R is a universal constant, R = 8:314 J/mol.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
827623
F*m = μ*Joules→ m = .25*40000000/1000 = 10000m = 10 km
188896
K = mv^2 = (10kg)(4.5 m/s)^2 = 202.5 J\nI would suggest you try doing your homework, asking for all the answers will only hurt you in the future.
201515
:}\nBeing a frozen chosen and loving happy clappers \nI'll let you know if I feel any increase in temperature
730160
The change in entropy is + 44.55 J/molK.\nThe enthalpy change is +242 kJ/mol, at standard conditions (SATP).  \nYou can use standard ambient temperature (298 K) for T.\n\n    Gibbs = +242.2 kJ/mol - (13.27 kJ/mol) \n               = +228.9 kJ/mol\n\nIf the temperature increased by more than a factor of about 20,  the second value would increase, and the Gibb's energy would be a negative quantitity.
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
704154
Joules (even though "s" is the last letter, this seemed to be the most reasonable guess)
801826
E=hf  where h = 6.6210^-34 Js\n\ne = 6.6210^-34*4.31*10^14 = 2.71 10^-19 Joule=1.7 eV
963007
multiply the mass of the water by the specific heat and the (temperature required minus the ambient temperature).that will give you the energy in joules.if you know in how much time you want this temperature to be achieved you can calculate the power as power is nothing but the rate at which the energy is dispensed
829976
You'll need this number:\n\n1.740×10^17 W.\n\nThis is the solar constant, the amount of energy received at the earth per second per area.\n\nAnd this number:\n\nsigma=5.67x10^-7 J/s/m^2/K^4\n\nThis is the Stefan Boltzmann constant which goes into the (surprise) Stefan Boltzmann law:\n\nJ=e*sigma*T^4\n\nSo, J is the energy received by the body per second per area, and T is it is temperature. The e stands for the efficiency of the body which is the ratio between the power it absorbs and then reemits.\n\nSo, you have an albedo which is an indication of how much light an object reflects from its surface versus how much it absorbs. With these numbers, you should be able to piece together your information with the equation above to get an answer. Let me know if you have further questions.
554942
Q =  m C (T-t)\nQ = (2kg)(2100j/kg)(0°c-(-20°C))\n= 84000 j\nthen\nQ = mL\n= (2)(335,000)\n= 670000 j\n\nthen\nQ = m c (T-t)\n= (2)(4200)(100-0)°C\n= 840000\n\nthen\n\nQ = mL\n= (2)(2,260,000)\n= 4520000\n\nthen\nTotAl = 8400+670000+840000+4520000\n\n\n= 6,038,400 J
736683
.065kg * 35 * specific heat of Aluminum\n\njust make sure your units check out; you may need to convert this equation into calories as specific heat is normally defined in joules I think.
362368
Given:\nm <mass of substance> = 12.0g or 12/1000 kg\nc <specific heat capacity of water> = 4180J per kg per degree celsius\nt <change in temperature> = 20 degrees to 100 degrees i.e. 80 degree celsius\n\nFind:\nHeat req.\n\nSolution:\nHeat required or Q\nQ = m * c * t\n  = 12/1000 * 4180 * 80\n  = 12/10 * 418 * 8 (cancelling 0s of 4180&80)\n  = 12/5 * 209 * 8 (div. 418&10 by 2)\n  = 12/5 * 1672 (mult. 207&8)\n  = 20064/5 (mult. 12&1672)\n  = 4012.8\n\nAnswer = 4012.8 Joules or 4.0128 KJ
512100
Well, it's 335 Joules.\n\n1 Joule = 0.238845896627496 Calories, so 335 Joules = 80 calories.
323553
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
701562
Well this is just like sound measurement. First of all you can calculate the amount of energy released from the eq. That would be equal to 2000000*(pi*48000^2)J/s = 1.45 x 10^16 J/s\ntherefore at a distance of 1km, it would be 1.45 x 10^16 J / (pi*1000^2)sm^2 = 4.61 x 10^9 J/sm^2. \n\nRate of energy passing through 5km^2 would be (4.61 x 10^9)x(5x1000)^2 = 1.152 x 10^17 Watts = 115200 Tera Watts