Idealized Heat Problem?
819964
On the vernal equinox, the sun is directly overhead the Earth's equator at local noon, and we are told that the insolation under those conditions is 1368 W/m^2.\n\nAt latitudes other than the equator, the solar energy per unit area is decreased by a factor of cos(L), where L is the latitude.  This is called the "projection effect".  At 40 degrees latitude, the incoming solar energy per unit area is therefore:\n\n1368 W/m^2 * cos(40 deg) = 1368 W/m^2 * 0.766 = 1048 W/m^2.\n\nI'm assuming the question is asking how long it would take to bring the water to a boil, not how long it would take to completely boil away the water.\n\nIn either case, you are missing a key bit of information necessary to answer the question:  the initial temperature of the water.  Let's assume that temperature is T.  I'll also assume that you are located as sea level, so the boiling point of the water is 100 C.\n\nThe specific heat capacity of water is 4.1855 J/(g*degC), and the density of water is ~1.00 gm/cm^3, so to bring one liter of water from temperature T to 100 C requires:\n\nE = 4.1855 J/(g*degC) * 1000 cm^3 * 1 gm/cm^3 * (100 degC - T) \nE = 4.1855*10^3 J/degC  * (100 degC - T)\n\nWe have a collector with a 1 m^2 surface area, so it supplies 1048 W = 1048 J/sec at 40 degrees latitude, all of which we are to assume goes into heating the water.\n\nWith these assumptions, it would take:\n\nt = [4.1855*10^3 J/degC  * (100 degC - T)]/(1048 J/sec)\n\nseconds to bring the water to a boil.\n\nFor instance, if T = 25 C, t = 299.54 sec ~= 5 minutes to bring the water to a boil.
1037424
evenly distributed energy
856026
Use the gas law\nPV=nRT
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.
810276
Well, probably easier to explain empirically -- I know that one of the equations that you would be looking for is Planck's law of black-body radiation.  That, along with other equations involved in transport phenomena could be used to model the system and let you know how much heat you lose from the glass.\n\nOff the top of my head, I can't remember every equation that one would want to use.  Good luck!\n\nThe link is to Wiki's entry for Transport Phenomena -- the first reference there would be an excellent -- if very heavy on the math -- reference for solving this problem.
775809
I believe the units of your supposed answer are wrong.  They should be kJ/mol and not kJ.\n\nHeat = m * Cp * DeltaT = 60 * 4.186 * (13.2 - 27.2) = - 3516 J\n\nThis heat is absorbed by 28.7 g of KI.  Per mole (MW(KI) = 166 g/mol), the heat will then be:\n\n28.7 g ===> 3516 J\n166 g ===> x\n\nx = 3516 * 166 / 28.7 = 20,340 J/mol = 20.3 kJ/mol
921632
PV = NRT, solve for N, note that T is absolute temperature in kelvins
163823
Q=mc(Tf- Ti)\nWhere Q is the heat added\nm is the mass\nc is the specific heat capacity\nTf is the final temperature\nTi is the initial temperature
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!
1003847
The study of heat and how it behaves.
977303
maybe what your teacher is trying to say here is the body temperature. now search for that answer. it's your homework.
575536
Here is a great article with sample problems.\n\nHope this helps!
196178
It is exothermic when the rxn. Heat is negative\nConsider Thermite\n\n2Al(s) + Fe2(O3)(s) –> Al2(O3)(s) + 2Fe(l) ∆Hºrxn=-847.6kJ
1017997
The equations are wrongor insufficient. Post the correct data if you want me to help/ Alternately, you could also message me.
956325
false - temperature is a measure of heat
409423
You have to specify the energy migration profile. Either adiabatic (zero energy transfer) or iso-volumetric (a lot of energy entering the system) or something in between.\n\nAlso, you have to specify the thermal units (Celsius/Fahrenheit/Kelvin/Rankin).
826242
rubbing 2 things together to create heat.\nRub yor hands together really fast and see what happens.