how much heat does it take to raise a gallon of water one degree?
115707
It depends on the pressure and temperature of the water so the previous answer is limited to only one condition.
208553
It should have lost 150 cal.\n\n1 cal per 1g water to raise 1º\n\nDrop of 15ºC multiply by the mass... 10g\n\n150 cal lost
100455
Here's the thing: \nAdding salt to water actually increases the boiling point of water, so depending on how much salt you use, your boiling point of salt water could be something like 105 degrees celsius instead of 100.  Boiling point raising is an example of a colligative property, which is a property that is not dependent on what kind of particles you're using, just how many particles there are.  So if you added more salt, the water's boiling point would be higher.  \n\nThat means that your water would have to be at a higher temperature to boil.  This means that reasonably, it would take longer for you to get your water to that higher temperature, but not by a lot.  It would be roughly the same amount of time.  \n\nThe trick, though, is that because your water is boiling at a higher temperature, whatever you're cooking in that water is going to cook faster.  You're cooking it at 105 degrees instead of 100, so whatever you're cooking in the water will be done faster.  That's probably what your high school chemistry teacher meant.
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?
973050
A calorie is the ammount of energy needed to raise the temp. of one ml of water one degree C.  A Calorie is 1000 times that, or the energy needed to raise one L. of water one degree C.
354143
increased pressure = increased heat
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
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
1057288
A Calorie is the energy to raise 1 k-gram of water 1 degree celsius, and there are 4.184 kJ per Calorie, so 12000 kJ equals 2868 Calories, which would heat 60 k-grams of water 47.8 degrees C.
141221
The original water temp is 25C, as was given in the problem... I'm assuming you meant to ask what was the initial temp of the water that was added, which isn't too hard. Here goes.\nQ(heat)= m(mass) x c(specific heat) x delta T(change in temp)\nso Q = (150g)(4.something J/gC-look it up)(25C-14C)\nBecause of conservation of energy and assuming ideal insulation, the heat LOST by the 150g of water during the mixing is equal to the heat GAINED by the 200g of water. So, once you have Q, rearrange the equation to look like [delta T] = Q/mc = Q/(200g)(c)\nThat will give you the temperature difference between 14C and the 200g of water's initial temp. Delta T should come out as being positive, since the temp is going up. \nWhen you have more complex mixtures of things with different specific heats and add processes and changes, the math becomes a lot more of a headache but the same basic principles apply just the same.
345993
Although adding salt to water will raise the boiling point, you will need to add a lot of salt.  If this is water that will be used for cooking anything, it will ruin it.  For example, to raise the boiling temperature 1/2 degree celsius, requires 1/4 Cup of Salt!\n\nWhen a pot is boiling over, just turn down the heat or remove it from heat.  You could add some salt, which will change the vapor pressure and slow down the boiling process, but it takes a lot.
1165215
Yes, this is called latent heat. The best example of this, is if you are boiling water at atmospheric pressure at sea level. If you placed a thermometer in this steadily heating water the temp would rise in a fairly uniform manner to about 96 degrees celsius, where upon it would seem to get stuck. The reason being, the heat energy is being used up to change the state of matter of the liquid water to steam. Until this transition is complete there will be no change in sensible heat as measured at the thermometer. Jules, lecturer. Australia.
1117920
It will be water at 100 degrees that will stay that temperature longer b/c of the increased volume.  The only way to increase the temp is to increase the energy (temperature) while keeping the same mass (like putting flame under the container of water).  Since you are increasing mass AND energy, it balances out.
168436
Water boils at 212 F.  The hotter the water, the less energy is needed to achieve this temp.  Of course pressure counts too.  Higher altitudes require different amounts of energy.  I think this is Doyles law.  Chem 101 would be a good class to learn this type of thing.  Also the purity of the water makes a difference.  The less pure the water (minreals, other impurities with lessen the amount of energy needed to achieve boiling temp.
680540
Assuming the ice is at 0 degrees, it would take 3.33 kJ (10 grams is 0.556 moles).  If the ice is not at 0 degrees, it will take more energy to raise its temperature.
621304
If you are dealing with a reaction involving ammonium carbonate in solution, and are then measuring the temperature rise, you need the heat capacity of water, not ammonium carbonate!\n\nPlease post the whole problem if that is not enough to get you started.
222680
temp 30-34 deg C\nadequate Nitrogen & Phosphorus\nwell watered but well drained\nNot too Humid\nC02 level 1000ppm