How to calculate heat loss?
149603
The amount of heat transfered is\n\n1) directlly proportional to the area of the sheet of given material. (A)\n\n2)directly proportional to the temperature difference on both sides of the sheet. (T2-T1)\n\n3)dirctly proportional to the time.(t)\n\n4)inversely proportional to the thickness of the sheet. (d)\n               \n---------------Q= A * (T2-T1) * t  /  d------------------------
615198
use the formula q=m*s*(t2-t1) where q is energy in heat,m is mass in grams,s is specific heat in joules/gram celsius,t1 is initial temp,t2 is final temp.\nsubstituting in formula,\n42,000=900*4.186*(t2-35)\nt2 works out to be 46.14 celsius.\nhope this helps.
741934
This question is not very difficult, but you have to be careful with the unit.\n\nFirst, to solve this question, you have to make the assumption that there is no energy lost to the surrounding. So, this reduces to a two system question: one system releases energy while the other one absorbs energy. In this particular case, the energy released by one system equals to the energy absorbs by another system.\n\nTo picture this, consider the following, one system goes through the following steps:\n\n[100 degC H2O (g)] --- (step 1) ---> [100 degC H2O (l)] --- (step 2) ----> [79 deg C H2O (l)]\n\nThe other system goes through:\n\n[260g, 19 degC H2O (l)] --- (step 3) ----> [260g, 79 degC H2O (l)]\n\nIt is obvious that both the energy involved in step 1 and 2 is negative (release energy) while the energy involved in step 3 is positive (absorb energy). We will now solve this question backwards. Starts from finding out the energy required by step 3, then find out the mass of steam that is needed to do the job.\n\nEnergy absorbed during step 3:\nE3 = m * c * (delta T) = 260 * 4.18 * (79-19) = 65208 J\n\nEnergy released in step 1:\n(you have to first convert heat of vaporization to unit of J/g: 40.7*1000/18 = 2261.111 J/g)\nE1 = m' * l = m' * 2261.111 = 2261.111m'\n(since we don't know how much steam is needed, let's just leave it as a variable)\n\nEnergy released in step 2:\n(the temperature of the same mass as above will be lowered to 79 degC)\nE2 = m' * 4.18 * (100-79) = 87.78m'\n\nTotal energy released = (2261.111 + 87.78)m' = 2348.89m'\n\nEnergy released = energy absorbed\n2348.89m' = 65208\nm' = 27.76 g
132159
Open up your thermodynamics book and do your own homework.\nHint.. add the energy required to raise the temp from 100C to 112C plus the energy used to change state, plus the energy used to go from 76C to 100C.  You should have all the info you need in your problem, and the energy required to do the change is in your book.
749307
amount of heat lost by iron bar = heat gained by water\n500x0.449x(212-T) = 2000x4.18(T-24)\nT is the equilibrium temp for bare and water.\nrearrange the equality and you will get T = 28.9 C\nso, the temp difference = 28.9 - 24 = 4.9 C.
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
174163
thermodynamics?  heat transfer?  physics?
755161
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.
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.
891159
use the equation\nE(enthalpy change) = m(mass) c(specific heat) DeltaT(change in temperature)\n\ntemperature change is initial temperature minus final\nsince temperature increased then the temp change is negative\n\n-2.00C
986703
In most physics homework problems, energy is conserved, and the following is true:\n\nEo = Ef\n\nHowever, in this problem, energy is lost due to friction.  Therefore...:\n\nEo - Q = Ef\n\nWe know that the child had potential energy at the top, and kinetic energy at the bottom.  Use this information to calculate the energy lost due to friction.  The energy lost due to friction will be denoted "Q" (Q usually stands for heat, which IS thermal energy):\n\nEo - Q = Ef\n\nRearrange the expression, since we're solving for F:\n\nQ = Eo - Ef\n\nNow plug in the formulas for potential and kinetic energy:\n\nQ = mgh - (1/2)mv²\n\nYou don't have to do this, but I factored out the "m" (mass):\n\nQ = m[gh - (1/2)v²]\n\nYou can plug things in now, since the values are all given:\n\nQ = 19.6[(9.80)(3.80) - (1/2)(2.52)²]\n\nQ = 668 J
501191
There is a formula...but this is much easier\n\nhttp://www.onlineconversion.com/temperature.htm
737811
You use the formal \nH= mcdeltat\n\nmanipulate it to be...\n\nc (specific heat capactiy) = H /(mdelta t)\n\nChange temperature to Kelvin,....4degrees = 277Kelvin\n\nC=221/(55X277)\n\n\nC= 221/15235\n\nC=0.0145 (in JOULES)
1140416
you need to employ th ebournoulli equation and the fanninf (or darcy ) friction factor equation ... \n\nThe Darcy-Weisbach Equation for pressure loss due to friction (in the SI metric system) is:\n\nΔP = 4f(L/D)(ρ v2)/(2 gc)\n\nwhere: \nΔP = pressure loss in Pascals \nf = Fanning friction factor (dimensionless) from Moody's diagram \nL = pipe length in meters \nD = pipe diameter in meters \nρ = fluid density in kilograms per cubic meter \nv = fluid velocity in meters per second \ngc = gravitational conversion constant = 1 (kg-m)/(N-s2) in the SI metric system\n\nAnd for head loss due to friction, the equation is:\n\nΔh = f(L/D)(v2)/(2 g)\n\nwhere: \nΔh = head loss in meters \nf = friction factor (dimensionless) \nL = pipe length in meters \nD = pipe diameter in meters \nv = fluid velocity in meters per second \ng = local gravitational acceleration of 9.807 m/s2
949133
OMG!\ni have the same question!\nmine is:\nA piece of an unknown metal has a mass of 20.0 g and at 100.0 C is placed in a styrofoam cup calorimeter containing 50.7 g of water at 22 C. If the final temp is 27.2 C, calculate the specific heat capacity of the metal. Assume that all the energy lost by the metal is gained by the water.\n\nok, haha fine, not the SAME question, but close enough. But, i'm uber confused. ><
773086
w = -(P x deltaV). Delta V is final V minus inital V (which you get by using the PV=nRT equation at each temperature). so -(3 x -11.8) = +35.39.\n\ndelta E = q + w. So add the q you got and the w you get.\n\ndelta H = delta E + (P delta V). \n\nThere ya go. hope this helps
929275
You would need the specific heat of the material.  The rate of cooling would be the total energy loss (E*A) divided by the specific heat.  This will be non-linear, and your value would be good for only the specific temperature given.\n\nWatts = Energy/time; one watt = 1 joule/sec.