of the metal and water is 20C, what is the specific heat capacity of the metal. I have tried hard and even though I am good at Physics, I still think I need help! Thank You
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Q=m.c.(delta)T\nQ=0.4 x 4200 x (20-16) = 6720 J\n\n6720=0.2 x c x (100-20) \nc= 420 J / kg K
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Lead — Specific Heat Capacity: 129 J/(kg*K)\n \n((129 J) / (kg * K)) * (.72 kg) * (7 K) = 650.16 joules
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So lets roll...\n\nmathematically, set it up as follows:\n\nLet x = the number of kg of water @ 82degC that must be added\n\nEnergy = m * c * T\n(m = mass, c = specific heat of water, T = tempearature)\n\nNeglecting convective and radiation heat losses:\n\n (x kg)*c*(82 C) + (0.2kg)*c*(14 C) = (x + 0.2 kg) *c*(36 C)\n\ndividing by c and re-arranging:\n\n82x + 2.8 = 36x + 7.2\n\nx = 0.096 kg
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All decent explanations so far; I hope I can add to what has already been posted.\n\nLiquid and Ice most definitely can exist at the same temperature and Ice can be and often does get colder than 0C.\n \nFirst of all the Triple Point of water is the pressure and temperature at which Ice, Liquid Water and Water Vapor exist at the same time in equilibrium.  Vapor is the right word, not steam although you can argue that the only time that Water Vapor and Steam are the same thing is at a very low pressure.  The Triple point of water only happens at a near perfect vacuum, 0.0063 ATM to be precise.\n\nAdding Salt to water will lower the freezing point, but then it isn't water anymore, it's salt water.  Even so, Liquid Salt Water and Ice Salt Water can also exist at the same place and time at the same temperature, slightly below 0C.  But let's stick to water.\n\nIce can easily be lower than 0 degrees.  Consider a bucket of water, liquid water sitting outside and the air temperature is -10C.  Little by Little, as the liquid reaches 0.0098 C, liquid starts to become ice.  The ice crystals form in the bucket, around the walls of the bucket and on the surface first.  The air is still -10C so the bucket, the liquid and the ice will keep getting colder.  The ice on the surface will drop to slightly below 0C but very slowly since it is still absorbing the heat of the not-yet-frozen water.  Eventually, the entire bucket of water will be frozen and will be just below 0C.  First the walls of the bucket will reach -10C (same as the air tempertaure).  The ice near the walls of the bucket and on the surface will reach -10C first, continually absorbing the heat energy from the relatively warmer ice near the center of the bucket.  Until finally, the entire bucket of ice is steady at -10C.  That is until the sun comes out or the day gets warmer.\n\n\nAnd by the Way, dry ice is not water, it's frozen CO2 (Carbon Dioxide)\n\nOne more thing, at a near perfect vacuum, you can actually cool liquid water to less than 0C without it freezing.  Never tried that but I did accidently freeze a beer last night...
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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.
548727
Considering water and aluminum as one system,\nthe energy balance gives delta U = 0;\n\nTherefore\n\nMwater *Cvwater* (T2 - T1,w) + Maluminim* Cvaluminum* (T2 - T1,a) = 0, where T2  = 50oC, T1,w = 120oC, T1,a = 20oC, Mi = Mass of component i, and Cvi = heat capacity of component i at constant volume.\n\nWith the given information, and looking for the heat capacities in a table, you can solve for Mwater.
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The heat capacity of glass (Pyrex) is 0.75 kJ/(kg*°C)\nThe heat capacity of mercury is 0.14 kJ/(kg*°C)\n\nYour heat energy needed to heat 1 kg of glass from 40°C to 50°C is 1 kg * 10°C * 0.75 kJ/(kg*°C) = 7.5 kJ\n\nYour heat energy needed to heat 7.5 kg of glass from 20°C to 50°C is 7.5 kg * 30°C * 0.14 kJ/(kg*°C) = 31.5 kJ\n\nIt will take more energy to heat the mercury.
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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)
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You could start with these sites and go from there.\n\nhttp://www.sciencebyjones.com/specific_heat1.htm\nhttp://en.wikipedia.org/wiki/Specific_heat_capacity\nhttp://www.rwc.uc.edu/koehler/biophys.2ed/heat.html
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Ha ha :))\n\n\nlet's say x be the length of metal rod 1 in the final composite rod \n\n===> (31-x) is the length of matal 2 in composite rod \n\nalpha 1(coefficient of linear expansion for metal 1 )  = change in length /original length  *change in temp =6.35* 10.^-2/31*100\n=\n\nSimilary ,for metal 2\n\n\nalpha 2  = (3.20 * 10.^-2)/31*100\n\nFOR THE FINAL compsite  rod  \n\nx*alpha1 * change in temperature + (31-x)* alpha 2* change in temp = 5.55 * 10.^-2 \n\nchange in temp above eqn is same =100\n\n0.002x+(31-x)*0.0010=5.55*10.^-2\n\n0.001x=0.0245\n\nx= 24.500 cm of metal rod 1\n\n& (31-24.500) cm = 6.500 cm of metal rod 2\n\n\nTHE author below has made a mistake of leaving the orriginal length :\n\nThe eqn is:\n\ndl=alpha*original length*dT
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You must approach this question with an understanding of energy conservation. Change in Energy = mass*heat capacity*change in temperature, where heat capacity = 1.00 for pure water. By evaluating your conditions with the above definition of thermal energy, it will show that you must supply 2 calories of heat to the system.\n\nIf you were to add another substance to the original system. You would once again approach the problem using energy conservation. The energy of a hot liquid (mass1*heat capacity*change in temperature) + the energy of a cold liquid (mass2*heat capacity*change in temperature) = mass(1+2)*heat capacity*change in temperature.\n\n-Christopher
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Thw formular to use here is Q = mcT where Q = quantity of heat, m = mass, c = specific heat capacity, and T is the change in the temperature, thus here Q = 2.0g*(30-25)*0.90 = 9J. Notice that I sed Joules as the unit.\n\n2.For this one, a law which i cant remember the name right now is nedeed. Here we have C(s)+O2(g)=CO2 Enthalphy -390KJ\nand also Mn(s)+O2(g)=MnO2(s) Enthalphy = -520KJ\nFor this law to be appplied, we have to write the second equation the other way to become MnO2= Mn+O2 Enthalphy = 520KJ. Notice that i changed the sign when i reversed the equation. Then O2 could be canceled out on the left side of the first equation and on the right side of the second equation. This was we are adding. Finally adding, we get MnO2(s)+C(s) = Mn + CO2 Enthalphy = -390KJ+520KJ = 130KJ.\nand thats the correct answer
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This question would have been much easier if the specific heat capacity of coal were known!\n\nNevertheless, let us model the coal to be anthracite of composition 85% graphite and rest 15% water. \nSo the s.h.c = 0.85*0.17 + 0.15*1 = 0.2945 cal/g-°C\nwhich gives heat required as 0.2945*(220-20)*5000 = 294500 cal\n=294500*4.184=1,232,188 J\nSo it seems the answer is C\n\nTo check this we can back calculate the s.h.c(specific heat capacity) using ans C which gives the graphite content in coal as 78.8% (quite realistic)\nUsing ans D the s.h.c is greater than for water(not physically possible) \nAnd A and B can be eliminated simply by order of magnitude analysis.\n\nIt seems most likely that C is the ans.
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What will work for you depends a lot on what the material to be coated is, on the application, time spent at temperature, heat cycling, peak temperatures, etc.\n\nHere's a non-ablative coating for metals that can withstand sustained temperatures of 850ºC (1560ºF) and thermal shocks from 1030ºC (1885ºF) immediately to a cold water bath:\n\nhttp://www.indestructible.co.uk/cat2/PL134-R1.html
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To solve this problem, one must know 4 peices of information.\nOne needs to know,\n-The specific heat of steam\n-The heat of vaporization of water\n-The specific heat of water\n-The heat of fusion of water\n\nOne would use these two formulas,\nQ = m*C*(delta T)\nwhere W is the heat energy, m is the mass, C is the specific heat, and (delta T) is the change in temperature.\nand\nQ = m*L\nwhere Q is again the heat energy, m is the still the mass, and L is the latent heat of fusion.vaporization of the substance.\n\nThis problem has 4 smaller parts one needs to solve in order to find the final answer.\nAs the steam is condensed into ice, 4 things happen in this order,\n-First the 110 degree C steam is cooled to 100 degrees C\n-100 degree C steam is condensed into 100 degree C liquid water\n-100 degree C liquid water is cooled to 0 degree liquid water\n-0 degree C liquid water is frozen into 0 degree C ice.\n\nFind the energy removed from the substance (H2O in whatever physical phase is appropraite at the time) and sum it up.
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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
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Latent heat of ice is 334kJ/kg.\nThis means, to convert 1kg of ice at 0c to 1kg of water at 0c it would require 334kJ (or 334000 Joules). \nSo, to convert 0.3kg of ice to water at the same temperature would require 0.3*334 or about 100kJ.\nThe answer would be (A) above.