Specific Heat of a Metallic Alloy?
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341668
Water evaporates at 100 degrees, celsius, so the liquid water you touched wouldn't be any hotter than that. The metal, on the other hand, does not. So it would be as hot as the object which is heating it.
437503
Q = m.s.ΔT\nHeat loss by the copper = heat gained by the metal.\nCopper (heat loss) = 200.0g x 0.38J/g.ºC x (195.5ºC-228.4ºC) = -2500.4J\ndivide the heat by the mass and ΔT of the metal to get the specific heat of the metal:\n2500.4J/(65.0g x (195.5ºC-18.3ºC)) = 0.2171 = 0.21 J/g.ºC (correct SF)
75209
I'll answer it, but you'd best pick me for the best answer if I'm doing all this work. ;)\n\nConvert C to K:\n\n178.0C = 451.15K\n25C = 298.15K\n30.6 = 303.75K\n\nTotal energy of water initially: Ei\n\nEi = 185.0g * 298.15K * 4.184J/g*K = 2.308E5 J\n\nTotal energy of water, final: Ef\n\nEf = 185.0g * 303.75g * 4.184J/g*K = 2.315E5 J\n\nEf - Ei = 700J\n\nLet Q be the specific heat of the alloy\n\nQ * 36.0g * 451.15K = Q * 36.0g * 303.75K + 700J\n\nQ*{(36.0g * 451.15K)-(36.0g * 303.75K)} = 700J\nQ = 700J/{(36.0g * 451.15K)-(36.0g * 303.75K)} \nQ = 700J/(16241g*K - 10935g*K)\nQ = 700J/3506g*K\nQ = 0.200 J/g*K
368284
Traditionally, metals have certain characteristic physical properties: they are usually shiny (they have "lustre"), have a high density, are ductile and malleable, usually have a high melting point, are usually hard, and conduct electricity and heat well. However this is mainly because the low density, soft, low melting point metals happen to be reactive, and we rarely encounter them in their elemental, metallic form. Metals are also sonorous, which means that they conduct sound well.\n\nAnd not all metals are magnetic!\n\nPalladium is a soft silver-white metal that resembles platinum. It doesn't tarnish in air. Palladium is the least dense and has the lowest melting point of the platinum group metals. It is soft and ductile when annealed and greatly increases its strength and hardness when it is cold-worked. Palladium is chemically attacked by sulfuric and nitric acid but dissolves slowly in hydrochloric acid. This metal also does not react with oxygen at normal temperatures.
331051
I listed some expansion coefficients\n\n\nMetal Temperature Range \n(oF) Thermal Expansion \n(microinch/(in oF)) \nAdmiralty Brass 68 - 572 11.2 \nAluminum  68 - 212 13.1 \nAluminum Bronze 68 - 572 9.0 \nBeryllium Copper 68 - 212 9.3 \nCast Iron, gray 32 - 212 5.8 \nCopper 68 - 572 9.8 \nCupronickel 68 - 572 9.0 \nHastelloy C 70 - 200 5.3 \nInconel 68 - 212 6.4 \nIncoloy 32 - 212 8.0 \nIron, nodular pearlitic 68 - 212 6.5 \nIron, pure 68 - 212 6.8 \nManganese Bronze 68 - 572 11.8 \nMonel 32 - 212 7.8 \nNickel Wrought 77 - 212 7.4 \nRed Brass 68 - 572 10.4 \nTin  32 - 212 12.8 \nTitanium 68 - 200 4.8 \nWrought Carbon Steel 70 - 800 7.8 \nYellow Brass 68 - 572 11.3
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The coefficient of thermal expansion for aluminum is 23*10^-6/ºK.  The expansion is ∆L/L, so ∆L/L = 23*10^-6*∆T\n∆Q = m*cp*∆T, where cp is the specific heat of aluminum = .897*J/(gr*ºK)\n\n∆T = 14500/(350*.897)\n\nTherefore ∆L = 20*23*10^-6*14500/(350*.897) cm
758041
brass is an alloy of copper and zinc, unless we have its composition, we cant tell its specific heat.
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Your answer depends on the nature of your problem, so you'll want to give more info if you want a complete answer.  The specific heat capacity you use depends on which substance is being heated up or cooled off.  The mass will be the mass of the substance that is being heated up or cooled off.\n\nOne exception that frequently occurs is when you are dissolving a substance in water.  Most problems will assume that the solution has the density and specific heat capacity of pure water.  So in other words, if you dissolve 3 grams of ammonium chloride in 100 mL of water and watch the temperature go down, most instructors would want you to use a mass of 100 grams (instead of 103) and the specific heat capacity of water 1.00 cal/g C or 4.184 J/g C if you're working in joules.\n\nHope this helps...this is the best I can do without more info about the problem.
705010
The specific heat of water is 4.184 J/g C.\n\nSince water's density is 1 g/ mL, 500 mL of water will have a mass of 500 g.\n\nThe equation to calculate the joules of energy required is:\nheat = mass x specific heat x change in temperature\n\nenergy = 500 (4.184)(95-25)\nenergy = 146440 J or 146.44 kJ\n\nHope that helped.
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Heavy Metal! Just joking. http://www.scienceproject.com/projects/intro/elementary/EX057.asp Go here, they have a lot of information.\n\nhttp://www.science.ca/askascientist/viewquestion.php?qID=1216 And this one. It is clear that aluminum conducts the best heat.
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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.
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Easy.  \n\nYou have q = mass * specific heat * change in temp.\n\nFind the specific heat of water and your set.  Q is joules so you need to convert.
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All Metals are tend to loose their electrons easily.\nSo if U heat an metal Rod then it's electrons gets energy by heating and start transferring to the other end of the Metal rod. So the other end is seemed to hot
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Aluminum: 900 J/kg/K\nCopper: 380 J/kg/K\n\nSo, yes. Aluminum has a higher specific heat and therefore a higher "amount of heat [is] required to change a unit mass of [aluminum] by one unit of temperature [than copper]." Why? Since specific heat is related to the excitation of the atoms and how well they hold heat, I would guess aluminum atoms bind to each other tighter and resists an increase in atomic motion (which would appear to us as an increase in temperature).
876892
Pressure, volume/specific volume, temperature, entropy, enthalpy, specific heat, atomic weight, and thats all I can think of off the top of my head
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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)