how much energy is required to bring 750g of water from room temperature (18 degC) to boiling point? (specific heat capacity of water: 4200J/kg?k)
204826
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)
641428
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.
409631
The specific heat of ice is .5 cal/g-oC\n--> http://www.school-for-champions.com/science/heat_ice_steam.htm\nThe ice changed temperature by 30 degrees\n\n445 g * 30 degrees *  .5 cal/g-oC =116.5 calories
509764
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
20713
1) We need to raise the temperature of the ice from -16.73°C to 0°C, which is a temperature rise of 16.73°C.\n\nMass of ice * specific heat of ice * temperature rise\n38.80g * 2.1 J/g K * 16.73° = 1,292.89 J\n\n\n2) We now need to add heat until all the ice is melted:\n\nMass of ice * Heat of Fusion\n36.80g * 333J/g = 12,254.40 J\n\n\n3) Now that the ice is all melted, it's all water, but the temperature is still 0°C. We now have to raise the temperature of the water to 100°C, which is a temperature rise of 100°.\n\nMass of water * specific heat of water * temperature rise\n36.80g * 4.2 J/g K * 100° = 15,456.00 J\n\n\n4) Now we have to change all the water into steam.\n\nMass of water * Heat of Vaporization\n36.80g * 2260J/g = 83,168.00 J\n\n\n5) All the water is now steam, but the temperature of the steam is still 100°C. We now have to raise the temperature of the steam to 227.6°C, which is a temperature rise of 127.6°.\n\nMass of steam * specific heat of steam * temperature rise\n36.80g * 2.0 J/g K * 127.6° = 9,391.36 J\n\n\n\nNow add up all the Joules\n1,292.89 J + 12,254.40 J + 15,456.00 J + 83,168.00 J\n + 9,391.36 J = 121,562.65 J\n\nConvert to kJ = 121.56kJ
98454
Okay, so molar heat capacity differs from normal heat capacity in the simple fact we're talking about a mole.  So water has a usual heat capacity of 4.186 J/(g*K).  All you have to do is multiply by the molar mass of water (18.0152.)  That should give you a molar heat capacity of 75.41 J/(mol*K)
174357
There are multiple steps here\n(1) Ice from -25 to ice at 0\n(2) Ice at 0 to water at 0\n(3)water at o to water at 100 (l)\n(4) water at 100 (liq) to water at 100 (vap\n(5) water  from 100 to 121\n\n(1) H1 = m x S x t\n(2) H2= m x l (latent heat of fusion)\n(3) H3= m x sx t\n(4)H4= m x L (latent heat of vapourization)\n(5) H5 = m x s x t     (all "t " corresponds to change in temp)\n\n(1) H1=70 x 2 x 25=3500\n(2) H2= 70 x 6.01x1000 =420700\n(3)H3= 70x 4.2 x 100 = 29400\n(4)H4 = 70 x 40.7x 1000 = 2849000\n(5) H5= 70 x 2 x 21 = 2940\n\nTotal Heat energy= 3305540 j/mol = 3305 KJ/Mol\n\nThis is the solution... I have a question: You gave the values of Phase change enthalpie in KJ/Mol... If they are correct , then this anser is perfect but if they are in J/mol then the answer u will get around 39KJ/mol\n\nThis solution helped you?
5447
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.
469078
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
532405
The heat capacity of something tells you how much energy you need to add to that something in order to raise the temperature by one degree.  In this problem, we know a temperature change, and we know (or can look up) the heat capacities of the things undergoing the temperature change.  That means we can calculate how much energy was added to the system (in this case by the energy corresponding to the enthalpy change of a reaction.\n\nLet C_s be the heat capacity of the sample, and C_c be the heat capacity of the calorimeter. Then the total heat capacity of the system is simply C_s + C_c.  If the temperature change is delta-T, then the energy change is given by:\n\ndelta-E = delta-T * (C_s + C_c)\n\nFor a process carried out at constant pressure, the change in energy is equal to the change in enthalpy, so delta-E = delta-H\n\nThe specific heat of water is 4.1855 J/(gm * K) (see source), and its density is ~1 gm/cm^3.  The combined volume of solution in the calorimeter is 100 mL, and we are told to assume that the density and specific heat of this solution is the same as pure water.  The heat capacity of the solution is therefore:\n\n100 cm^3 * 1 gm/cm^3 * 4.1855 J/(gm* K)= 418.55 J/K\n\nThe total heat capacity of the system is:\n\n(418.55 + 12.1) J/K = 430.65 J/K\n\nThe temperature change was 6.1 degrees C = 6.1 K, so\n\ndelta-E = delta-H = 430.65 J/K * 6.1 K = 2627 J\n\nNow, this enthalpy change was for the reaction of 50 mL of a 1M CuSO4 solution with 50 mL of 2M KOH solution.  50 mL of 1M CuSO4 solution only contains 0.05 liters * 1 mole/liter = 0.05 moles of CuSO4.  If we want the enthalpy change PER MOLE of CuSO4 reacted, we have to divide the enthalpy we just calculated by 0.05:\n\ndelta-H (J/mole CuSO4) = (2627 J)/(0.05 mol CuSO4) = 5.254*10^4 J/(mole CuSO4)
29685
You need to check your units.  The specific heat of water is not 4.184 J/g; it is 4.184 J/g*degrees C (or degrees kelvin).  And the heat of solution should not be in J/degree, but in J/g.\n\nTry this equation instead:\n\n[HoS = J/g] = [(Specific Heat = 4.184 J/g*C)*(mass = 100g)*(Temperature change = Initial C - Final C)] / [Mass of CaCl2 = 3 g]\n\nBy subtracting the final from initial you should get the negative value and not need the -1 at the front.\n\nSometimes enthalpy of solution is in J/mol rather than J/g, in which case you modify the above by calculating the mols of CaCl2 in 3 grams and switching that out with the 3g in the above equation:\n\n(3g) / (110.99 g/mol) = 0.027 mol\n\nTheoretically, you should want to know the heat capacity of the CaCl2 because it is also being heated, but this is assumed to be negligible.  Write that into your error analysis and you teacher will love you.  Also, make sure to check your significant figures; i don't know how many sig figs you used in weighing out your 3 grams, but that will be necessary to know how much to round everything.  Good luck.
573929
Assume the kettle is 2500watts Thats about 25p per hour. It takes about 4 minutes to boil, so 60 divided by 4 is 15 hence  25p divided by 15 =1.7p to boil a kettle
251622
The average specific heat capacity of a human body is approximately 3500 J/kgK
537961
The heat given off by the burning reaction (delta E) is equal to the heat gained by the calorimeter.  (m * change in temp)\n\nYou can calculate the heat released by burning.\n.553 g with a molecular wt. of 88g/mole \n(.553 grams) / (88 (grams / mole)) = 0.00628409091 moles\n(975 (kJ / mole)) * 0.00628409091 moles = 6 126.98864 joules\n\nYou can calculate the heat absorbed by the water (3000 g * .356)\n3 000 * .356 = 1 068 joules\n\nyou can calculate the difference, it must have heated up the apparatus.\n6 127 - 1 068 = 5 059 joules \n\n5059 joules = x grams * .356\n5059 joules / .356 = 14210.67 grams of water\n\nThe metal apparatus was had a heat capacity equivalent to 14210.67 grams of water.
431293
Conservation of energy.\n\nCalculate the amount of heat stored in the 28ºC water using 0ºC as the starting point.  This energy must first melt the ice, for which you'll need to use the heat of fusion of water.  The remainder of the energy will warm up the entire amount of water.
650830
Water acts as the medium in many processes. 25 deg C is the room temp.
291413
The latent heat of fusion -- once you cool water to 0°C, you have to remove more heat to actually crystallize it.\n\nIce at 0°C requires the addition of heat to melt into water at 0°C.  That heat comes from whatever you are cooling.\n\nThe latent heat of fusion for ice is 333.6 joules/gram