calculate the heat loss of a cooler?
542138
The general formula to calculate the heat loss is\n\nQ=m*cv*(t2-t1)\n\nQ is the heat loss\ncv is the specific volume at constant volume(if it is a constant volume process)otherwise use cp(specific heat at constant pressure)\n\nt2 is the external temperature\nt1 is the internal temperature
395354
generally cooling of the mass depends on mass surface , heat conductivity coefficient of material which is losses heat and temperature difference between cooling mass and media. \nif there is a liquid  then evaporation , humidity of media , temperature and pressure of media , kind of media eetc. also important\n\nmain heat losses = K . (T2-T1) . A \n\nK = coeeficient of heat conductivity ( depend material , thickness ) \nT2 - T1 = temperature difference \nA = surface area
366921
Q = mc delta T\nHeat = mass times specific heat times change in temperature\nspecific heat for water is 4.186 kJ/kg C\n-41840 J (negative because absorbed) = .200 kg x 4.186 x 10^3 J/ kg C x (Tf - 20)\nTf = final Temp.\nsolving the equation gives you Tf = -30.0 degrees Celsius
911974
Go with a line graph, that way you can also find out (or estimate) temperatures at times when you didnt measure it.
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
672712
start to worry when it gets up to 150 or so degrees fahrenhite.\n\ni think that chip will run around 100 - 120 degrees with a "normal" hsf and cooling setup
75209
I'll answer it, but you'd best pick me for the best answer if I'm doing all this work. ;)\n\nConvert C to K:\n\n178.0C = 451.15K\n25C = 298.15K\n30.6 = 303.75K\n\nTotal energy of water initially: Ei\n\nEi = 185.0g * 298.15K * 4.184J/g*K = 2.308E5 J\n\nTotal energy of water, final: Ef\n\nEf = 185.0g * 303.75g * 4.184J/g*K = 2.315E5 J\n\nEf - Ei = 700J\n\nLet Q be the specific heat of the alloy\n\nQ * 36.0g * 451.15K = Q * 36.0g * 303.75K + 700J\n\nQ*{(36.0g * 451.15K)-(36.0g * 303.75K)} = 700J\nQ = 700J/{(36.0g * 451.15K)-(36.0g * 303.75K)} \nQ = 700J/(16241g*K - 10935g*K)\nQ = 700J/3506g*K\nQ = 0.200 J/g*K
174163
thermodynamics?  heat transfer?  physics?
422010
Its easy...\nHeat lost by meat = heat gain by contents in calorimeter(ie water)\n\nm X s X delta(t) = M X SX Delta(T)\nm-mass of lead                  M-mass of water\ns-specific heat of lead         S-Specific heat of Water\nt=change in temp=(373-x)              t-Change in Temp = (303-x) because you dont know final temerature\ns of lead= 0.032cal/g (i think)  S of water= 1cal/g\n\nsubstitute these values .. you will  get 53 degrees C as final answer\n\nI hope you got the point.\nGood luck\n\nNote: there may be some calculation mistakes, but the theory is the same
141221
The original water temp is 25C, as was given in the problem... I'm assuming you meant to ask what was the initial temp of the water that was added, which isn't too hard. Here goes.\nQ(heat)= m(mass) x c(specific heat) x delta T(change in temp)\nso Q = (150g)(4.something J/gC-look it up)(25C-14C)\nBecause of conservation of energy and assuming ideal insulation, the heat LOST by the 150g of water during the mixing is equal to the heat GAINED by the 200g of water. So, once you have Q, rearrange the equation to look like [delta T] = Q/mc = Q/(200g)(c)\nThat will give you the temperature difference between 14C and the 200g of water's initial temp. Delta T should come out as being positive, since the temp is going up. \nWhen you have more complex mixtures of things with different specific heats and add processes and changes, the math becomes a lot more of a headache but the same basic principles apply just the same.
437503
Q = m.s.ΔT\nHeat loss by the copper = heat gained by the metal.\nCopper (heat loss) = 200.0g x 0.38J/g.ºC x (195.5ºC-228.4ºC) = -2500.4J\ndivide the heat by the mass and ΔT of the metal to get the specific heat of the metal:\n2500.4J/(65.0g x (195.5ºC-18.3ºC)) = 0.2171 = 0.21 J/g.ºC (correct SF)
157547
The maximum possible coefficient of performance, B, can only happen in an ideal heat pump, or the Carnot heat pump. The formula for coefficient of performance is:\n\nB=Tl/(Th-Tl) where Tl is the low temperature and Th is the high temperature, both in kelvin.\n\nB= 266 K/(295 K - 266 K)\nB=9.17 is the maximum possible coefficient of performance.
867420
about heating maching bring a thick wire and wrap it around a spoon then the tips connect them into electric source (big one) dont do it alone it gets really hot ...ok\nabout the cooling machine i dont know
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!
742159
So lets roll...\n\nmathematically, set it up as follows:\n\nLet x = the number of kg of water @ 82degC that must be added\n\nEnergy = m * c * T\n(m = mass, c = specific heat of water, T = tempearature)\n\nNeglecting convective and radiation heat losses:\n\n (x kg)*c*(82 C) + (0.2kg)*c*(14 C) = (x + 0.2 kg) *c*(36 C)\n\ndividing by c and re-arranging:\n\n82x + 2.8 = 36x + 7.2\n\nx = 0.096 kg
958888
Use the formula:\nQ (heat) = Cp (Heat Capacity) * delta T (Change in Temperature)
897593
If you are doing this in terms of the system,\n\nsince the system is losing energy (delivers heat) then the amount of energy is "-1275 J".  The surrounding add heat (perform work on) to the system, so the amount of energy is "+855".  The internal energy of the system is the sum of the energy loss and the work done to the system, so -1275 J + 855 J = -420 J.\n\nIf you are doing this problem in terms of the surroundings, the all the signs are reversed and you should end up with +420 J.