in terms of physics and specific heat. for example, water is 4180j/kg.k etc.
207757
Hydrogen has the highest known specific heat of 1.43 x 10^4  J/kg*K at a temperature of 25°C.
233733
1 calorie heats 1g of water by 1 deg C.\n630cal heats 630g of water 1 deg C\n630cal heats (630/15)g water by 15deg
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
54244
the difference in the boiling point of this soln compared to pure water equals Kb times m, where Kb is the boiling point constant(0.512 for water) and m is the molality of the soln.\nso the answer si 100.512 degrees C.
353788
Well,  Hot & Cold are not only subjective (What's Hot for he is cold for me) and relative.  Better to think of temperature.\n\nTemperature is the measure of average kinetic energy.  The hotter, the more energy and vice versa.  \n\nAlso, as temperature is an average, some of the "hot" water may be cold.  In fact, there could be a several molecules that are frozen - we'd never know - but the average temperature is "hot."\n\nWhen you mix them heat is transfered from the high average kinetic energy water, "Hot," to the low average kinetic energy water, "Cold."  The "Cold" water absorbs the heat in the transfer and the results is as follows...\n\nThe "Hot" water is cooled, and the "Cold" water is warmed.  The results is 24 oz's of water that has an average kinetic energy, or temperature, higher than that of the "Cold" water, and lower than that of the "Hot."
527473
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. ><
390765
First the specific heat of water is not needed because both items mixed are water and would have the same specific heat so just use the weight.\n\nso you have 60 * 80 + 40 * 25 = 100 * x    where x is the new temperature in degrees C\n\nIf the items being mixed were different then use the specific heat of each item to calculate the number of joules in each item which will give the total number of joules in the combined item.
186689
I will take a stab at it.  I am guessing it is because water has more degrees of freedom.  In the liquid there is also rotational degrees of freedom to absorb energy that the solid ice does not have.\n\nTo Neil, the formulas are for heat capacity and have nothing to do with the way something is heated, only the amount of energy that it takes.\n\nNor does it have anything to do with Newton's law of cooling.
70304
use the fmla.... Q=msT\n\nwhere Q is the quantity of heat required, m is mass, s is specific heat capacity of water, T is the temperature difference. i dont hav a calculator with me now, u do the computations plz...
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.
417814
Quantity of energy lost by the water:\n\nQ = (m)(deltaT)(c)\n    = (50.0g)(2.00 C)(4.18 JK-1g-1)\n    = 418 J\n\nSince the water got colder, the reaction was endothermic, and the enthalpy change for the reaction is +418 J.
260006
Q =  m C (T-t)\nQ = (2kg)(2100j/kg)(0°c-(-20°C))\n= 84000 j\nthen\nQ = mL\n= (2)(335,000)\n= 670000 j\n\nthen\nQ = m c (T-t)\n= (2)(4200)(100-0)°C\n= 840000\n\nthen\n\nQ = mL\n= (2)(2,260,000)\n= 4520000\n\nthen\nTotAl = 8400+670000+840000+4520000\n\n\n= 6,038,400 J
521877
The Specific Gravity of water is 1
361700
Look at your units for latent heat. Joules per g. That means it takes 334 J to convert each gram of ice to liquid water. You have 250 g of ice. That should tell you what the formula is.
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
236447
Well... break it down to formulas. 1 calorie is the energy needed to heat 1 gram of water by 1 degree celcius.\n\nSo to heat 1000 grams of water (1 liter) by 1 degree, it takes 1000 calories. Now double the water to 2 liters and it will take 2000 calories.\n\nIf you are adding heat at a rate of 1000 calories per minute, then 1 liter of water heats at 1 degree celcius per minute and 2 liters heat at a rate of 1/2 degree celcius per minute. So yes. It is linear.\n\nHOWEVER (there is always a however) there are many factors that go into this in a lab setup. Namely, a larger volume of water will have more surface area and disipate heat to the suroundings faster. Also, a larger pan (wider) will heat faster than a taller pan. Etc, etc, etc.
401588
In thermochemistry, in many cases, the heat (either being absorbed or released) can be determined using the general equation\n\nE = m*c*(delta T)\n\nE is the energy, has the unit of J\ndelta T is the change in temperature, has a unit of K, determined by final temperature - initial temperature\nc is the specific heat capacity, usually has unit of J/g K\nm is the mass, has the unit of g\n\nHOWEVER, if you use the specific heat capacity that has a unit of J/mol K, m will have the unit in mol. So, you have to be consistent.\n\nUsing c in J/g K,\n\nE = m*c*(delta T)\n= 108 * 0.88 * (660 - 25) = 60350.4 J = 60.35 kJ\n\nUsing c in J/mol K to solve,\n\nE = (108/26.98) * 23.7 * (660-25) = 60242.62 J = 60.24 kJ\n\n**note that the small difference is due to the possible round off error in the specific heat capacity.