How much heat energy is required to raise the temperture of 5 kilograms of coal from 20 ' C to 220' C?
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What is the specific heat of coal (carbon)?\n\nQ = mc(deta T)\nQ is the heat energy, m is the mass, c is the specific heat and delta T is the change in temperature.
951441
Coal is actually very inefficient as an energy source.  One ton (2000 lbs or 1000 kg +/-) of coal produces the same amount of heat energy as 18,000 cubic feet of natural gas, 160 gallons (1100 lbs) of oil, or 5 ounces of uranium.  Also coal produces much more waste than any of the other energy sources.
454506
100 degrees celcius\n212 degrees Fahrenheit
232210
heat coal in an oxygen free atmosphere. without oxygen it will not burn but bond of the carbon and hydrogen atoms will be broken.\n\nCheck out an aquarium store.
166194
About 350 watts heat is to be dissipated from an area of about 0.5 meter square and limiting surface temperature below 100 degree celsius.
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.
870267
.4084c(80)+.4084(heat of vaporization)\nc is specific heat of water\nthat will give you J/KG C   si units
800141
The question is if you start with gas at 20.0 atm and 20.3 C and heat it to 125 C what will be the pressure?  It is a changing condition problem making use of Charles' Law.  The answer is 27.1 atm.
737043
The answer is (e).\n\nTo calculate the enthalpy change, find the product of the specific heat of water, the mass of the water in grams and the temperature change of the water in kelvins. Since the change in degrees celcius equals the change in kelvins, use (-2).\n\nE (enthalpy change) = 4.18*200*(-2) = -1678 J \n\nI hope this helps!
14789
A general form of the equation I beleive you are looking for comes from a basic understanding of thermodynamics.  I am assuming that your ct=(Tfinal-Tinitial)...\n\nSo a general form of the equation would be:\nmCp(Tf-Ti)=Q \n\n"Cp is the specific heat"\n\nWATCH UNITS!!!!!  \n(I am ignoring the conversions in this statement)\n55g*Cp(67degrees UNITS)=343 cal \n\nSolving this equation will give you Cp ... just make sure that you watch units.
817267
I think such questions bug you because you equate Internal energy with heat. Work and heat both are energies in transit for any process. Think like this, if 220 J work is done on the system its energy would have increased by 220 J.  But instead its internal energy has decreased by 500 J. So only way this is possible is if [220 + 500] J = 720 J amount of energy leaves the system as heat.  If you like to work with formula then\n\nU = Q - W\n\nhere U = -500; Q unknown: W = -220\n\n-500 = Q - (-220)\n-500 = Q + 220\n\nor Q = -500-220 =  -720 J\n\nSo the system has lost 720 J as heat
870952
To answer this Question you need to use the Conservation of Energy Law.   This takes its simplest form in the idea that:\n\nHeat Lost by the Steam    =   Heat Gained by the cold water\n\nYou will need to say:\nLet L be the unknown Quantity and form an Equation:\n\nHeat Lost by the Steam:\nThe steam (one assumes at 100Celsius) condenses to water at 100Celsius.  In doing this it gives up its Latent Heat of Vaporization (mL)   The water it forms is then cooled to the final temperature giving up this Internal Energy to the System.  The Internal Energy(Heat) gained is given by m.s.dT where dT is the change in temperature.\nNote that we have two  Energies to Add Together here.\n\n\nHeat Gained by the cold water:\nThe Internal Energy(Heat) gained is given by m.s.dt where dt is the change in temperature.\nBoth the cold water and the Calorimeter are heated from their starting temperature(15C) to the final temperature(70.6C)\nNote that we have two  Energies to Add Together here.\n\nNow write down the equation:\nHeat Lost by the Steam (steam + hot water)     =   Heat Gained by the cold water (cold water + calorimeter)\n              \nm1.L + m1.s.dT  =  m2.s2.dt + m3.s3.dt\nAnd solve for the Unknown L
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Get an electrical heating element, a DC power supply, a thermometer, a voltmeter and an ammeter.\nPass electricity through the heating element, dipped into a measured volume of water (or salt water) in an insulated cup, and measure the current, the voltage and the time taken. This will give you the heat energy supplied. Measure the temperature rise, and work out how much energy would be required to raise 1g of water (or salt water) by 1 degree Celcius. The answer should be close to 4.2!
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Energy required to heat 1liter of water:\n\nQ = V(liters) * d(kg/liter) * Cp(kJ/kg/K) * (Tfinal - Tinitial)\nwhere Cp is the specific heat capacity of water, which is temperature dependant but for which you can use an average value for the temperature range.\n\nQ = 1 * 1 * 4.18 * 73\nQ = 305.14 kJ\n\n1 kW.h = 3600 kJ\n\nQ = 305.14 kJ / 3600 (kJ/kW.h)\nQ = 0.085 kW.h\n\nPrice = 0.085 * 10 = 0.85 cents
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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.
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.
1036179
I agree with 'keulenae' but the only exceptions are a cubic meter is approximately 35.3 cubic foot and a power plant has an efficiency less than 1 (at the best case reaches 64%). Therefore:\n1GWH=10^9WH=3.6x10^12 joules\nassuming the average efficiency of power plant be 50%:\nRequired input energy = 7.2x10^12 joules\nHaving each 5.2x10^7joules equal to thermal energy of one Kg gas:\nRequired Gas = (7.2x10^12)/(5.2x10^7) ~ 138461.5 Kg\nAssuming 0.7Kg = 1 cubic meter = 35.3 cubic foot we get:\nRequired Gas = 138461.5 x 35.3 / 0.7 = 6982415.6 cubic foot\n\nSo it burns about 7 million cubic feet to get 1GWH electricity!\n(Don't forget that the above gas volume is almost the normal DAILY production of an standard treating train in gas refineries).