What is the specific heat capacity of LNG?
480410
The specific heat of liquefied methane is 3.50 J/g/K. The specific heat capacity of LNG is likely to be close to this as LNG is about 90% methane. If you need a more precise value, you will need to find out how much methane, ethane etc is in the LNG and look up the appropriate specific heat values and calculate a weighted average. See page 2 of the URL below.
635964
Check this: http://www.eia.doe.gov/oiaf/analysispaper/global/index.html\nOther useful links:\n\nGazprom - http://www.gazprom.ru/eng/\nWikipedia - http://en.wikipedia.org/wiki/LNG
540230
For pentane which is close to gasoline it is .74 x 10^3 per degree F\n\nI can't find diesel but 35 deg API crude is .44 x 10^3\n\nBecause these are liquids, the term is cubical expansion i.e. incremental volume change for a degree rise in temperature.
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. ><
569971
This problem only has enough information if you assume it is an ideal gas.  In this case, PV=nRT is the equation of state.\n\nn and R are constants in this case, so PV/T must also be constant.  These values must also all be in absolute scales - the celcius must be converted to kelvin or any other scale based on absolute zero.  This means:\n\nP(i)V(i)/T(i)=P(f)V(f)/T(f)\nV(f)=T(f)P(i)V(i)/[P(f)T(i)]\n    =(250+273)K(740mmHg)(410mL)/[(680mmHg)(27+273)K]\n    =[523*740*410/(680*300)]mL\n    =777.83mL\n\nBy the significant digits you've presented, this can only be evaluated to any accuracy as 780mL.  (two significant digits)
190244
Specific Heat Capacity\nThe specific heat capacity of a solid or liquid is defined as the heat required to raise unit mass of substance by one degree of temperature. This can be stated by the following equation: \n\nQ= mc(changes of T)\nwhere, \nQ= Heat supplied to substance, \nm= Mass of the substance, \nc= Specific heat capacity, \nT= Temperature rise.
842595
You measure the amount of fuel put in.  Using the flow of that fuel (cubic feet per hour), you then determine the amount of energy in each cubic foot of that fuel (or look it up - Natural gas is 1020 btu/cubic foot).  Then measure the amount of heat given off and the amount of electric produced.  \n\nEnergy out (Btu's) / Energy in (Btu's) = efficiency.  \n\nYou need to convert the electric out to btu's by multiplying the kW to Btu's (multiply by 3412).  With the heat given off, you need to determine the temperature of the exhaust and the amount of air flow of that temperature and that will give you the thermal side of the output.
1118780
Your question is very open.\n\nheat equations very general can be:\napplied heat = specific heat x mass x increase of temperature\napplied heat = heat capacity x increase of temperature\napplied heat = mass x specific melt heat\napplied heat = mass x specific evaporation heat (at boiling point)\nHeat from a supply = power x time.\nCombinations and others are possible.\n\nwave equation can be x = A sin( 2pi t/T)\nThere are more possibilities.
705505
The specific heat of water is 1 calorie/gram °C = 4.186 joule/gram-°C.  If you want to use 4.18, that is okay.\n\nThe heat added = the specific heat of water (c) times the mass of the water (m) times the change in temperature.\n\nQ = cmT or\n\n= 4.18 joule/g-°C * 125 g * 2.44 °C\n= 4.18 * 125 * 2.44 joules (you can do the math)\n\nThis is for 1.75 g of CaCl2 which has a molecular mass of 111.0 g/mol, so the number of moles is:\n\n1.75 g / 111.0 g/mol = 0.01577 moles of CaCl2.\n\nDivide the heat change (in joules) above by 0.01577 moles to get the heat per mole.
84471
Higher specific heat capacity liquids can absorb more heat energy per unit of liquid than lower specific heat liquids.\n\nIn most applications the higher specific heat liquid will be the liquid of choice as it can "carry away" more heat per unit volume.
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).
454337
750g = 0.75 kg\nDelta T = 100-18 = 82°C (or K)\n\nEnergy required = 0.75 x 82 x 4200 = 258300 Joules\n\n258.3 kJ\n(also 0.07175 kWhr, or 244.8 BTU)
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
868501
That depends on the specific heat capacity of the oil and the amount of oil in the drum. A rough calculation could solve this assuming the drum is insulated and the amount of water is small. \nEnergy required=mass of oil (kgs)*specific heat capacity of oil (J/kg/degrees)*70 (degrees celcius)\n\nI have used 70 here assuming 20 degrees celcius room temp and assuming you are talking in celcius also. If not, you must convert 90 fahrenheit to celcius and take away room temp.\n\nNow convert 30 minutes to seconds; 30*60 = 900 seconds\nSo you want the energy required to be delivered within 900 seconds, therefore the power you need in watts is the energy required divided by 900.\n\nThis heater power will work in ideal conditions but should be increased (maybe by a factor of 2) to account for assumptions.\n\nI don't know how oil heating works (Don't blame me if you explode!!!) or what "heater capacity" means but I hope the answer in watts will help you.
131645
You want "energy density:"\n\nEthanol (Ethyl Alcohol):\n26.8 MJ / kg\n= 2905 kCal / lb.\n\nCrude Oil:\n45 MJ / kg\n= 4879 kCal / lb\n\nAviation Kerosene:\n43 MJ / kg\n= 4662 kCal / lb\n\nI don't understand the "If measured with a calorimeter, what degree celsius would it reach." Are you talking about the combustion temperature?
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.
358448
specific heat? that would be represented by Cp.\nHowever if you are using delta H then it would either be: delta H (fus) which is the enthalpy of fusion, used in problems concerning melting point (indicating a state change from solid to liquid)\nOr it could be delta H (vap) which is the enthalpy of vaporization, used in problems concerning a boiling point (indicating a state change from liquid to gas)\n\nHeat calculations are done using the formulas:\nq=m x (delta T) Cp <-- Used when there is a temperature change, but no state change\ng= m x (delta H fus) <-- Used for state changes from solid to liquid\nq= m x (delta h vap) <-- Used for state changes from liquid to gas