1) What is the energy change (in kJ) when the temperature of 20g of water increase by 10 degree celcius?
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2)\n\nAs deltaG=deltaH - TdeltaS\n\ndeltaH needs to be multiplied by 1000 to make it in J\n\nPut into the equation you get: T=279.26K\n\nie 6.1 degrees Celsius\n\nthe importance is that this is the temperature at which the reaction becomes thermodynamically spontaneous.  Basically the benzene will solidify at 6.1degrees.
797125
The heat lost by the warmer water will equal the heat gained by the cooler water.  The change in the heat will equal \nq = (s)(m)(T1 - T2)   where q is the heat change (units of Joules (J)), s is the specific heat of water (4.184 J/gK), m is the mass of the water in grams, and T1 and T2 are the initial and final temperatures of the water samples in Kelvin.  Assume the density of water = 1.00g/mL.\n\n(4.184J/gK)(30g)(Tf - 280K)  =  (4.184J/gK)(50g)(330K - Tf)\nTf is the final temperature of the combined solutions.\n\nSolving for Tf:\nTf = [50(330) + 30(280)] / 80  = 311K
914713
Just assume the energy gained is the potential gravitational energy U=mgh, with h=100 meters, and then the heat capacity of water, etc.
765294
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.
252583
Since 1 cal is the energy required to raise them temp of 1g of water 1º C... the temp would fall 1.1111111111111 etc. degrees C for each 1000 cals lost.
143592
Heat required to rise the temperature of  ice from \n-10 degree centigrade to 0 degree centigrade \n= ms(t2-t1)\n=10 * 2.1 * 10  g* {joule/(g*degree)} (degree)\n=210 joule\n\nHeat required to convert ice to water  \n=mass x specific latent heat fusion of ice \n=10 * 336  g*joules per gram\n=3360 joule\n\nHeat required to rise the temperature of  water from \n0 degree centigrade to 100 degree centigrade\n=ms(t2-t1)\n=10 * 4.2 * 100\n=4200 joule\n\nHeat required to convert water to steam\nmass x specific latent heat of vaporisation of water\n10 * 2264\n22640 joule\n \nHeat required to rise the temperature of steam from \n100 degree centigrade to 110 degree centigrade \n= ms(t2-t1)\n=10 * 2 *  (110-100)\n=200 joule.\n\ntotal of all is  30610 joule.
819865
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.
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The steam.  It would have to go through a phase change and give up a bunch of energy to turn back into water at 100.  Assuming you mean equal mass of water and steam. \n\nSame question for ice.  How much energy does it take to melt 1 Kg of ice at 0 to water at 0?
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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°).
1066111
let final temp be x\nHeat lost by 0.5gm h20 = mass*specific heat(a constant)*(98 -x)\n= 0.5 *4.2*98-x\n= 2.1(98-x)\nheat gained by 0.235kg water = 235*4.2*(x-20)\n=987(x-20)\nheat lost = heat gained\n2.1(98-x) = 987(x-20)\n205.8 - 2.1x = 987x - 19740\n989.1x  = 19945.8\nx = 20.165 degrees C\nfinal temp = 20.165 deg C
734516
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.
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.
341576
Besides the loss of heat due to the difference with room temperature, we can also check how water specific heat behaves with temperature. \nThe specific heat is the amount of heat required to change a unit mass of a substance by one degree in temperature. This specific heat increases with temperature, so this indicates us that water at a higher temperature needs a bit more energy to rise 1º than water at lower temperature.\n\nSo the answer is yes, it will take more time to rise from 90º to 100º because of two reasons:\n1- There is more loss of heat in the water at 90º than water at 10º.\n2- Specific heat is higher in water at 90º than water at 10º.
721040
I get 45.13 degree Celsius.\nReasoning:\nChanging 23 gram ice in water of 0 degree requires 7.66e3 Joule, according to part 1. This amount of energy taken from the 70 degree water decreases the temperature of the latter by 7.66e3 x (70-10) / 2.9e4 = 15.85 degrees (using part 1 and 2).\n\nSo now we have a mixture of 23 gram water at 0 degree and 115 gram (5 times as much!) at 54.15 degrees. This will average out to (5/6) x 54.15 = 45.13 degree Celcius.
948401
Energy required to heat 1liter of water:\n\nQ = V(liters) * d(kg/liter) * Cp(kJ/kg/K) * (Tfinal - Tinitial)\nwhere Cp is the specific heat capacity of water, which is temperature dependant but for which you can use an average value for the temperature range.\n\nQ = 1 * 1 * 4.18 * 73\nQ = 305.14 kJ\n\n1 kW.h = 3600 kJ\n\nQ = 305.14 kJ / 3600 (kJ/kW.h)\nQ = 0.085 kW.h\n\nPrice = 0.085 * 10 = 0.85 cents
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It's not, that is only a certain method of measuring a calorie.  A calorie is the amount of energy required to raise the temperature of 1 gram of water 1 degree celsius at 1 atmospheric pressure.  \n14.5 and 15.5 (celsius) are merely the temperatures chosen to start and end at.  There is also the 20 degree calorie (19.5-20.5) and 4 degree calorie (3.5-4.5). The differences in energy required from different temperatures vary in the thousandths of a joule.
753535
Just use the formula\n\nE = m*Cp*(Delta T)\n\nE=energy\nm=mass\nCp= Specific Heat\nDelta T = change in temperature\n\nCp= 0.51998 J/(C*g)\n\nGood Luck with the test!