how much heat would be needed to raise the temperature of 50g of water from 20 degrees to 25 degrees?
920577
actualy it depends on the pressure because the formula is:\nQ=m*Cp*Delta(T)\nAnd Cp for water is dependant on Temprature you are giving heat to the water. you can find the analytical method for finding Cp in Thermodynamics  van wylan ed 6
246402
Latent heat of vaporization of water is 2260 kJ.kg-1.\nConverting 50g to kg = 50/1000= 0.05kg\n\nTherefore 2260 KJ/KG x 0.05KG = 113 KJ or Kilojoules or 113,000 J or Joules is your answer
482741
You need a few different constants to figure this out.\n\nWater has a Specific Heat of 1 kcal/kg/°C\nThat means it needs 1 kcal per kilogram to heat the water 1 degree celsius\n\nIt has a density of 1000 kg/m3\n\nIt has a Latent heat of Fusion of 80 kcal/kg (melting) and 65 J/kg (freezing)\n\nSo if you know the volume you can work out the weight of the water. Once you have the weight (or you already know it), you can work out the rate of the drop (or increase) of water temperature.\n\nOnce you get to the point of freezing, or melting, simply add on the latent heat, and you can then get the total amount of energy required to change the water from one temperature to the next.\n\nSo for example;\n\n1) How much energy to heat 2 kg water from 18°C to 100°C is =>\n\n2kg * 1 * (100-18) = 164 kcal (The heat up curve is fairly straight)\n------------------------------->\n\n2) 2kg of water from 5°C to -10°C is as follows;\n\n(2 x 1 * (5 - -10)) + (65 * 2) = 160 kcal\n\nCurve drops down fairly constantly, then flats out until the water turns to ice, then continues to drop until you are at your temperature.\n\nI hope that helps a little.\n\nIf you want to figure out the times for all this to happen, you need to either know what heat source you are using (for example the wattage of the kettle element), and also you need to figure out the heat losses from the container (e.g. kettle or pipe, etc.)\n\nThese can all then be converted to an equivalent kcal number.
608826
It's in reference to density and is the ratio to the density of water at 25 degrees celcius.\n\nBasically, water is 1 g/cm^3. So something with a specific gravity of 25 will have a density of 25 g/cm^3.
904003
Temperature rise of water = 2 deg C = 275 K\nq=mc(change in temperature)\n  = (100)(4.18)(275)\n  =114950J\n  = 114.95 kJ\nHmm. Either I'm wrong or the question is incomplete. Did the question perhaps give more information on the reaction?
78350
When the metal enters the water, it begins losing heat to the water. At the same time the temperature of the water rises. This process continues until the temperature of the metal and the temperature of the water are equal st which point (30.6oC) no net flow of heat occurs. The heat lost by a gained system is given by (mass)(Specific heat)(change in temp.)First calculate the heat gained by the water, then calculate the specific heat of the metal.\n\nTemp rise in the water - \n   (change in temp) 30.6 - 25.0 = 4.4\n\nHeat gained by water - \n              \n   (185.0)    (4.184J)/(g oC)  (4.4) = 3.4*10 pwr3 J\n\nThe metal dropped into the water must have the same temp as the water (30.6)\n\nTemp drop by the metal - \n   (Change in temp) = 178-30.6 = 147.4\n           \n Specific heat = \n(heat)/(mass)(change in temp)\n\nSpecific heat of the metal =\n(3.4*10 pwr3)\n(36g)(147.4)\n             \n=0.640 J/g oC\n\nhope this helps. good luck
341668
Water evaporates at 100 degrees, celsius, so the liquid water you touched wouldn't be any hotter than that. The metal, on the other hand, does not. So it would be as hot as the object which is heating it.
177044
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...
274980
Since there is no temperature change in the substance, only a phase change (solid --> liquid), all you need to use is,\nQ = Lm\nwhere Q is the heat energy, L is the latent heat of fusion, and m is the mass of the substance.\nI dont happen to know the latent heat of fusion for water (I think it is like 2260 J/g, maybe), but I am sure it is in your book somewhere.\nJust plug the values into the formula and you are set.\n\nEDIT:\nThe Latent heat of fusion of water is 333 J/g.\nThe latent heat of vaporization is 2260 J/g.
93911
From ?. You have to know the starting temperature.\n\n By definition, 1 BTU will raise the temperature of 1 pound of water 1F degree.\n\nSo subtract the starting temperature from 18 and then multiply by 45.
223397
1) find specific heat of coal.  Coal is pure carbon; use that if they don't give a "coal" specific heat.\n\n2) That is the heat required to raise 1 GRAM of that substance by 1 degree celsius.\n\n3) 5000 g = 5 kg\n\n4) 220-20 = 200\n\nSo you get:\n(specific heat of coal) * 5000g * 200 = energy required.\n\nThat's all the help I'm giving on your homework ;)
222680
temp 30-34 deg C\nadequate Nitrogen & Phosphorus\nwell watered but well drained\nNot too Humid\nC02 level 1000ppm
186344
At 20 degrees C, the vapor pressure is 17.54 mmHg.  At 25 degrees C, it's 23.76 mmHg.  You'd get close if you interpolate between these two points, even though the curve, strictly speaking, is not linear.  \n\nOr, if you're ambitious, use the Antoine equation:  \n\nAntoine Equation Parameters\n\nlog10(P) = A − (B / (T + C))\n\n    P = vapor pressure (bar)\n    T = temperature (K)\n\nFor temps in the range 273. - 303. Kelvin, A = 5.40221, B = 1838.675 and C = -31.737.
869302
1) Calculate the heat needed to melt 100 grams of ice completely: \nMass of ice * Heat of Fusion\n100g * 333J/g = 33300 J\n\n2) Calculate the cooling we can obtain \nMass of water * specific heat of water * temperature loss\n180g * 4.2 J/g K * dT = 33300 J => dT = 44.0 degrees\n\nConclusion we need only to melt 100g * 30/44 = 68.1 gram of ice to cool down the water to zero degree. 31.9 grams of ice remains.\n\n3) 1) Calculate the heat needed to melt 50 grams of ice completely: \nMass of ice * Heat of Fusion\n50g * 333J/g = 16650 J\n\n4) Calculate the cooling we can obtain \nMass of water * specific heat of water * temperature loss\n180g * 4.2 J/g K * dT = 16650 J => dT = 22.0 degrees\n\n5) Calculate the average temperature of the water.\nAfter melting of the ice we have 50 g of water at 0C and 180 g of water at 30-22=8C. This average out to 6.3C
749307
amount of heat lost by iron bar = heat gained by water\n500x0.449x(212-T) = 2000x4.18(T-24)\nT is the equilibrium temp for bare and water.\nrearrange the equality and you will get T = 28.9 C\nso, the temp difference = 28.9 - 24 = 4.9 C.
128148
It would depend on the amounts (mass) of the water.\n\nedit:\nThe specific heat of water is not 1 [artless_man], it is 1 calorie per gram degress celcius, the units make all the difference.  Notice the lower case c in calorie.\nA "better" specific heat of water, using metric units, is 4.186 J / (gram * degree celsius).\n\nHowever, for this problem, you dont even need to know the specifc heats since both substances are water.
251863
At 20 degrees C, it's 0.6g/100mL. You can convert that to a Ksp.