I need to learn how to calculate energy changes from Liquid to gas. i dont understand the whole energy reuqired equals specific heat compacisty multiplied by mass of water multiplied by temperature change.
215628
see dear ,\nwhile calculating the energy changes we need to know the hidden heat { latent heat } that is absorbed during the change in state . for eg when we are calculating the heat absorbed or given out during a change in state , like ice to water , just first calculate the latent heat ie \nq=mL{ heat=mass *latent heat } now add this value to \nq=mct{heat=mass*specific heat*change in temp} keep in mind that specific heat is a constant value \nwater's specific heat is 4180j/kg/degree Celsius.
157181
law of conservation of energy\nany gas that is compressed will 'lose' its heat energy to the ambient temperature in the process (ie a refrigerator). in the specific example the gas (1 cm3 at 70F )is at room temperature until its expanded to lets say 1000 cm3, now 70F is spread out 1000x the area,(as is the gas in the can) the more the 70F is expanded the 'colder' it gets (much like the universe or a car tire, but its all relative)
286162
a.)\n\nIn a liquid or a gas, not all molecules have the same kinetic energy.  Temperature is just a measure of the average kinetic energy.  There is statistical range of kinetic energies.  Those molecules that have the higher kinetic energy may escape from the liquid into the gas phase.
159526
Q = mc delta T\nHeat = mass times specific heat times change in temperature\nspecific heat for water is 4.186 kJ/kg C\n-41840 J (negative because absorbed) = .200 kg x 4.186 x 10^3 J/ kg C x (Tf - 20)\nTf = final Temp.\nsolving the equation gives you Tf = -30.0 degrees Celsius
188106
Its easy...\nHeat lost by meat = heat gain by contents in calorimeter(ie water)\n\nm X s X delta(t) = M X SX Delta(T)\nm-mass of lead                  M-mass of water\ns-specific heat of lead         S-Specific heat of Water\nt=change in temp=(373-x)              t-Change in Temp = (303-x) because you dont know final temerature\ns of lead= 0.032cal/g (i think)  S of water= 1cal/g\n\nsubstitute these values .. you will  get 53 degrees C as final answer\n\nI hope you got the point.\nGood luck\n\nNote: there may be some calculation mistakes, but the theory is the same
368706
First calculate the number of moles of acid and base that react by multiplying the molarity by the liters of solution.  The total volume (mass) of water is the sum of the two volumes (masses).  The calories attributed to the water are calculated by the change in temperature times the mass of water times the specific heat of water.  The calorimeter soaks up 480 J/degree times the temperature change, which is pretty stout.  (I am separating the two heat capacities.  If the 480 figure is for the water plus the container, then just use 480 times the temperature change.)  Then scale the answer up to one mole of reactant.
248186
water is incompresible, so ...\n\nVgas = Vtank - Vwater\n\nm(gas) = M(gas)*Niu(gas)\n\nNiu(gas) - no. of moles \nM(gas) - molecular mass of gas\n\nNiu(gas) = m(gas)/M(gas)\n\nwe have the following equation:\n\nPV = Niu*R*T, where:\n\nP - pressure (N/m^2)\nV - volume (m^3)\nNiu - no. of moles of gas\nR - universal gas constant (8.31 in these conditions)\nT - temperature (K)\n\nP = Niu(gas)*R*T/V(gas)\n\nP = [m(gas)/M(gas)]*R*T/[Vtank - [m(water)/(water density)]]\n\nin formula introduce Vtank in m^3 (not in liters 1 m^3 = 1000 dm^3 = 1000 liters), and the pressure will result in N/m^2\n\ntransform it in psia ... and the probleme is solved
450955
You need to know the heat capacity for water (Cp) and its heat of vaporization (Hv).  Assuming Cp is given for moles, convert 141 g of H2O into moles.  Multiply the difference in temperatures (100 - 41) times Cp.  Multiply the value you get by the number of moles.  Now you have the amount of heat it takes to raise the water temp. from 41 to 100 deg.C.  Next, multiply the heat of vaporization (per mole) by the number of moles.  That's the heat needed to convert all of the water into vapor.  Add the two heat values together for the answer.\n\nNote:  Your question states "answer in kJ/mol."  If that's what they really want, just do the above for 1 mole; you wouldn't need to know the mass of the water.  But I think if they just wanted the answer in kJ/mol, they wouldn't have said there  was 171 g of H2O.  Also...while most references give Cp and Hv in heat per mole, some give it in heat per mass; so be careful.
388949
The change in heat = specific heat * mass * change in temperature (Q=C*g*ΔT).\n\nAssuming that the Aluminum also ends at 28.3 °C, then the change in heat for the Aluminum is\n\n0.0902 J/g°K * 35.7 g * (28.3°-81.9°)\n\nThe above will be negative as heat was lost from the Aluminum.  The change in heat for the water is\n\n4.184 J/g°K * 75.0 g * (28.3°-24.9°)\n\nThe above will be positive as the water gained heat.\n\nThe Heat Capacity of the calorimeter is the difference between the heat lost by the Aluminum and that gained by the water divided by the change in temperature (which is the same as for the water (28.3°-24.9°).
250069
Formula:  J = K*m*(T2-T1)\n\nSince both K & m are 1 for your problem, J = 125 Joules\n\nK is the specific heat defined as 1 for water.  Varies with other materials.  Also this formula is not strictly true for 1 liter unless you assume the 1 liter weighs 1 kg.  Liters are volume and the formula is for mass.
619164
First, find the energy required to get the ice to 0 C:\n\nmass of ice X (0 - temperature) X Heat capacity = Energy (ice)\n\nThen, find the energy required to melt the ice:\nconvert the enthalpy of fusion to units of J/g, (you will need to determine the molar mass of water) and calculate\n\nmass of ice X enthalpy of fusion = Energy (melt)\n\nAdd the energy from the ice and the energy of melting the ice together.  \n\nEnergy (ice) + Energy (melt) = Energy (cold)\n\nEnergy (cold) / (mass of water X heat capacity of water) = temperature drop from 87 C.
623726
The first part of the question is pretty easy.  Just figure out how much energy must be removed from the water to convert it to ice, then that's how much it puts into the room.\n\nHeat Energy out = 3.34x10^5 J/kg * 81.1 kg = 27,087 kJ\n\nBrad didn't understand the second part.  What they want to know is this:  Assuming you have a perfect Carnot engine, calculate the efficiency of that engine and use that to determine the extra energy coming into the system.  Use the equation:\n\nEff = W/QH = 1 - TC/TH\n\nwhere\nEff is Efficiency of the system (percent)\nW is the work done by the system (energy exiting the system as work) \nQH is the heat put into the system (energy entering the system) \nTC is the absolute temperature of the cold reservoir (deg kelvin)\nTH is the absolute temperature of the hot reservoir (deg kelvin)\n\nFirst, find the system efficiency using Th and Tc (converted to kelvin):\nEff = 1 - (273.15 / 298.25) = 0.084 (8.4%)\n\nNow we already figured out the heat ejected into the room.  W = 27,087 kJ.  Next, we solve for QH.\nQH = W/Eff = 27,087 / 0.084 = 321,865 kJ\n\nIn our case:\nQH = heat from ice + energy supplied to device, so\n\nenergy supplied to device = 321,865 kJ - 27,087 kJ = 294,778 kJ\n\n\nHad to edit this one.  It was nagging at me, and I knew it didn't seem right.  In my original answer, I was using the efficiency equations for a carnot heat engine.  I was wondering why engine efficiency should go up with an increase in Th.  It didn't make sense.  I should have used the equations for a refrigeration cycle.  My bad.  Here's the right answer:\n\nFor a refrigeration cycle, the heat ejected into the room is equal to the heat from the ice plus the extra work.  \n\nFirst thing, we figure out the heat removed from the water to make the ice:\n\nHeat from ice = Q = 3.34x10^5 J/kg * 81.1 kg = 27,087 kJ\n\nNext, calculate the Coefficient of performance:\n\nCOP = Q/W = TC/(TH-TC)\n\nwhere\nCOP is the Coefficient of performance (efficiency equivalent for refrigeration)\nW is the work added to the system (energy entering the system as work) \nQ is the heat put into the system (heat energy entering the system from the water to make ice) \nTC is the absolute temperature of the cold reservoir (deg kelvin)\nTH is the absolute temperature of the hot reservoir (deg kelvin)\n\nCOP = 273.15 / (298.25 - 273.15) = 10.88\n\nNow, we can find the work energy added to the system (answer for part 2):\n\nW = Q/COP = 27,087 / 10.88 = 2,490 kJ\n\nAnd then we know that the heat ejected into the room is equal to the heat from the ice, plus the work:\n\nHeat into room = Q + W =  27,087 + 2,490 = 29,577 kJ
371123
I can't answer the question, but you need Boyle's Law, the volume of a gas is inversely proportional to it's pressure.  Also, are you assuming there is no temperature change?
15915
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
327828
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.
25909
Note: I'm going to convert all the intermediate steps into kJ by dividing by 1000. I won't show that step.  \n\nFirst, let's find out how much heat it'll take to get the ice from -17.87 degrees to zero degrees, where it has to be in order to melt.  This is a change of 17.87 degrees.  (Because one degree Celsius and one Kelvin is the same size degree, they are interchangable.) We'll use the specific heat capacity of ice.  \n59.90 g ice * (2.1 J/g*K)*(17.87 K)=2248 J = 2.248 kJ\n\nNext, let's find out how much energy it would take to melt that ice at 0 degrees to water at 0 degrees using the heat of fusion:\n59.90 g ice * (333 J/g)=19950 J = 19.95 kJ\n\nNext, let's find out how much energy it would take to get the water from 0 to 100 degrees (so it can become steam) using the specific heat of water:\n59.90 g water * (4.2 J/g*K)*(100. K)=25160 J = 25.16 kJ\n\nNext, let's find out how much energy it would take to evaporate the water using the heat of vaporization: \n59.90 g water* (2260 J/g)=135400J = 135.4 kJ\n\nFinally,  let's find out how much energy it would take to get the steam from 100 to 250.7 degrees for a total change of 150.7; we will use the specific heat capacity of steam:\n59.90 g steam* (2.0 J/g*K)*(150.7 K)=18050 J= 18.05 kJ.  \n\nNow, just add up all of the heats required at each of these steps: \n2.248 kJ+19.95 kJ+25.16 kJ+135.4 kJ+18.05 kJ=200.8 kJ, with correct significant figures.
74026
The guy who answered before was right but i wanna explain it. An exothermice reaction is one in which heat energy is lost to the surroundings, thus at the end of the reaction, some heat would have been lost. For heat to have been lost, the product has to have lesser heat than the reactant.Take not that heat here is the same as Enthalphy.\nSince the Enthalphy heat value = heat of product-heat of reation = a negative number because heat of product is lesser.\nthus the answer is i nd ii\n\n2. To calculate the heat absorbed or evolved in a system, the formulais Q(quantity of heat) = m*c*T, where m is the mass, c is the specific heat capacity, and T is the change in temperature, thuis this three are neede. Density is not important in this calculation\nso the answwers are i and iii