how much energy is needed to heat up nichrome "red hot"?
117733
That is a function of\n\nH = I^2 R t {a quantity of heat generated by a current applied for a time across a resistance}\n\nH=Heat in joules {1 joule = .24 calories\nI=Current in amperes\nR=resistance in ohms\nt=time in seconds\n\nIf you wanted to make the equation a rate\nJ/s=I^2•R \n\nSo what you need to know essentially is the resistance of your wire and its heat dissipating capacity\nthen you can balance the impedance with heat loss\n\nAs for a cigarette lighter in a car look at the fuse size and multiply by 12 volts to get wattage {a rate}\n\nA thinner wire will have a greater resistance\nA longer wire will have a greater resistance\nand both will lead to greater heat loss through surface area\n\nThe best thing you could do is place a rheostat and an ammeter in series with your wire and experiment\n\nAlso remember that the resistance of your element will increase slightly as it grows hot this principle is called impedance
1128546
specific heat Q = mc(t2-t1)\nput everything in SI units..\nthen \n\n40,000= 500/1000 * c(68-20)\n40000=0.5* 48*c \nhence c= 40000/24= 1666.67 joules/kg oC.
675016
This is a calorimetry problem! There is an equation for it:       \nq = mc(change in temperature)\n\nq = the heat required \nm= mass\nc= specific heat \n\nAlso, write out the reaction and products.
48024
Your question doesn't make sense.  No metal would produce heat by being put in hot water--heat is only produced by a chemical change.  Perhaps sodium metal, which would conbust if the water had any oxygen in it, but it wouldn't need to even be hot water.  Also, water wouldn't be water at 400 degrees.
846365
Thermal energy IS heat.
249911
You wouldn't want a conductor for a heating coil, conductors aren't supposed to heat. You would want a resistor that would generate heat but still conduct the electricity needed. Nichrome Wire is generally considered the best and most common heating element.
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.
476810
http://ushotstuff.com/Heat.Scale.htm
137865
Deppends on the bulb, and it's efficiency, total energy in = total out. 50W at 10 min is 30000 Jouls of energy.\nThe efficency of an average incad. bulb is about 15% efficient, so I would say you get about 4500J of light, and the rest would mostly be heat, about 25500J of heat.\n\nI am not sure how to convert the Jouls into Lumens, and Calories/BTUs, but it can be done.
982154
calories=the energy needed to heat 1 gram of water 1 degree\nso to get the answer  you take 50 x 7 which gives you how many calories it takes
977909
HOT-IF YOU KNOW WHAT YOUR DOING!!
174163
thermodynamics?  heat transfer?  physics?
613796
how does an indoor firing range use energy?
961600
In the laws of thermodynamics, it is stated that energy can not be created nor destroyed, but may be converted from one form of energy to another.  So, it might be better to ask how does one convert various forms of energy into thermal energy?\n\nWell, in the world of HVAC, we can burn a hydrocarbon fuel such as oil or gas (energy), and the products of the combustion process will be heat energy, light, water vapor, carbon dioxide, carbon monoxide (we try to create very little of this stuff), and vaporous and particulate matter (smoke .. again, we try to create very little of this stuff too).  The heat and light energy can be placed next to, or mixed within substances that have a lower temperature, and thus heat energy will flow from the hotter products of combustion to areas and substances that are cooler (heat energy always travels from areas of higher temperature to areas of lower temperature).  If a heat exchanger is used, the above products of combustion will be on one side of the heat exchanger, and substances such as cool air or relatively cool water will be on the other side, heat energy will flow from the hot products of combustion to and through the cooler heat exchanger material, and then will flow from the heat exchanger to the even cooler substance your trying to heat such as the water or air.\n\nElectrical energy can be converted to heat energy (thermal energy) by passing a current through a device that has resistance (a resistor).  It takes electrical pressure (voltage) to push the electrons through a resistive material, but any flow of electrons through the resistive material will cause the material to heat up.  The formula P = I x E applies here and “P” will be the rate of power measured in watts, “I” stands for flow of electrons measured in amps, and “E” is electrical pressure measured in volts.  To convert watt hours to BTU's (a watt hour can be a watt that is running for one full hour), you multiply watt hours times 3.413 (I have seen conversion factors ranging from 3.412 to 3.415, I’m not sure what the exact conversion is, but any number in this range will be pretty accurate).\n\nThere are other ways of converting other forms of energy to heat energy, including mechanical energy (friction), other chemical reactions, solar radiation, nuclear reactions  etc.
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. ><
622160
which ever ones' hotter
1165181
It's been awhile since i've done this but i'll give it my best\n\nFirst off, the first energy you need to calculate is the energy needed to raise the temperature of ice to 0 degrees C. so...\n\n(12.37 g)(.48 cal/deg/g)(35 deg) = +207.816 cal\n\nWhen the ice changes to water you have a positive heat of fusion. So...\n\n(12.37 g)(80 cal/g) = 989.6 cal\n\nNext you have to raise the water temperature from 0 degrees C to 100 degrees C. So...\n\n(12.37 g)(1.0 cal/deg/g)(100 deg) = 1237 cal\n\nNext you have the positive heat of vaporization.\n\n(12.37 g)(540 cal/g) = 6679.9 cal\n\nNext once it is steam, you must raise the temperature from 100 degrees C to 146 degrees C. So...\n\n(12.37g)(.48 cal/deg/g)(46 deg) = 273.1296 cal.\n\nNow that you have the energy for each step add them together.\n\n207.816 + 989.6 + 1237 + 6679.9 + 273.1296 = 9387.45 cal\n\nI'm pretty confident in my breakdown but it defintely wouldn't hurt to have someone else check it