What am I doing wrong in this Heat problem?
608010
epsilon=0.6\nsigma=5.6704*10^-8 [J/(s*m^2*K^4)]\nP=Q/A=120 [W/m^2]\n\nWe know:\n\nP=epsilon*sigma*T^4\n\n120=0.6*5.6704*10^-8*T^4\n\nSolving for T we get:\n\nT=(120/(0.6*5.6704*10^-8))^(0.25) = 243.7 [K]\n\nThen, using Newtons cooling law:\n\nT(t)=T(environment)+(T(initial)-T(environment))*e^(-rt)\nT(t)=293+(243.7-293)*e^(-r*t)\nT(t)=293+(-49.3)*e^(-r*t)\n\nNow, I'm stuck, because I don't know the rate of change of temperature (r).  Maybe you try a different formula!?
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
1078838
I don't know, but mine do the same. \ngood idea why don't you or I invent the heated\nmouse? I could use the bucks.
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.
745748
Impossible to answer. How cold does it get? How high do you have the heat set for? How old is the burner? How efficient is it? Where do you live? Too many variables.
791970
Use the Gay-Lussac gas law: V/T is constant. Note that T here is the temperature in Kelvin, not in Celsius. Luckily 27 C = 300 K
71035
Not 100% sure. But I think its because the air inside isn't moving. The air inside is stationary, while it is continuously circulating outside, which causes changes in the temperature. IMHO
640958
1degree Fahrenheit of heat +1degree Fahrenheit of heat =1degree Fahrenheit of heat bkz there is no heat transfer
1119303
sounds like an overheating problem.....when you changed the processor did you use the old heatsink/fan? is the fan working properly? try changing the heatsink/fan and use a good heat transfer paste like artic silver 5
726134
How long?  In time?  Like, seconds? \nI guess it depends on the intensity of the heat source.
753535
Just use the formula\n\nE = m*Cp*(Delta T)\n\nE=energy\nm=mass\nCp= Specific Heat\nDelta T = change in temperature\n\nCp= 0.51998 J/(C*g)\n\nGood Luck with the test!
961642
The initial mass of the steam is the unknown you're supposed to solve for.  That's why you're not given it!
294886
check out event log for finding the exact problem. it's greatly exprected that the problem is due to overheating.
1136926
A simple solution is to mulitply the celsius temperature by 2 and then add 31 (32 - 1). It's not the exact science, but It will give you the right answer 97% of the time.
263946
Have each page work on the same princable as the bi-metallic thermometer.  Try to find two materials the expand differently than eachother because of heat, or you could use humidity to expand one material more than the other. You could probably find some material that contracts (or expands) when an electrical current passes through it. place a strip of this along the top and bottom of each page. Thats all i can think of right now.\nCool question by the way.
811304
let thetemperature of the mix be 't'*Cfind the heat lost by coffee\nfind the heat gained by ice in three stages\na)from -2.2*C to 0*C\nb)for melting and becoming water at 0*C\nc)for getting to the equilibrium temperature of the mixture\nequate and solve
522868
Take it to someone who knows what they are doing. \n\nWe've answered this question about 8 times for you. \n\nBring it to me, and I'll fix it for you - only if you quit posting the same question.