More water-temperature questions?
537035
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.
742159
So lets roll...\n\nmathematically, set it up as follows:\n\nLet x = the number of kg of water @ 82degC that must be added\n\nEnergy = m * c * T\n(m = mass, c = specific heat of water, T = tempearature)\n\nNeglecting convective and radiation heat losses:\n\n (x kg)*c*(82 C) + (0.2kg)*c*(14 C) = (x + 0.2 kg) *c*(36 C)\n\ndividing by c and re-arranging:\n\n82x + 2.8 = 36x + 7.2\n\nx = 0.096 kg
426622
temperature differences
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.
511312
See my question about the age of the Appalachian Mountains.  And the similarities/ differences w/ Himalayans.  Oh yeah, Does hot water freeze faster than cold water?  If so, why?
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.
56978
water, water and more water
471637
1. 1080 degrees\n\n2. 1800 degrees
947952
Highlight the data and put your cursor at the bottom right of the lowset box, the cursor should turn into a small +. hold the cursor down and drag your mouse doun a few rows.  Once you let up on your mouse you should see the info you want.\n\nYes, if you increase pressure you can increase the temperature of the water, but it needs to be in a closed system, it's called super heating.
977303
maybe what your teacher is trying to say here is the body temperature. now search for that answer. it's your homework.
617601
This is not as simple of a question as you have assumed.  The rate of cooling will depend on the mass of the water, that is, how much water you have.  Secondly it will depend on how well the container conducts heat.  Thirdly, what is the surface area of the container.  The greater the surface area, the faster the water will cool.  You must also consider air movement in the immediate area of the container.  If there is air movement, the cooling rate will be affected by how much air flows over the surface of the container.  Next, you must know the temperature at which the water boils.  The higher the altitude the lower the boiling point.  Finally what is the temperature and humidity in the room?
562654
Okay, I'll try this one -\n\nRun this experiment twice (or more times), each time with a piece of metal having different density.  The metal piece should be of adequate size (a sewing needle isn't going to do it), it needs to be big enough to significantly affect the temperature of the water.  You will need to know the mass of the metal (i.e., weigh it) and have a thermometer capable of reading from room temperature up through boiling or so.  Please note this is not going to give you a highly accurate answer, but can give you a ballpark idea (I'm guessing you're probably not doing this in a 500 million dollar lab).\n\nThe procedure:\n\n1)  Measure and record room temperature as T1.\n\n2)  Weigh the metal, call it M1.\n\n3)  Bring a known quantity of water (weigh it, or measure it and convert to weight, call it M2) to a temperature just short of boiling (say 180F to 200F, use the same temperature for each run of this experiment).  Record the temperature as T2, and remove the heat source (take it off the burner, etc.).\n\n4)  Drop in a metal piece at room temperature.  Allow a couple of minutes for heat to exchange.  Make sure you use \nthe same amount of time for each run of this experiment.\n\n5)  Measure and record the new temperature of the water as T3.\n\n6)  Repeat with your other metal samples.\n\nYou can now estimate the specific heat capacity for each metal.  It will not be super-accurate (heat escapes your system, and temperatures won't have completely equalized), but should give you some idea.  Just assume all temperatures did equalize and that no heat escaped for calculational purposes.\n\nSince (you are assuming) the heat which left the water went into the metal, then the product of mass, specific heat capacity, and temperature change should be the same for both the metal and the water (but in opposite directions).\n\nThis means that:\n\n(M1)(c1)(T3-T1) = (M2)(cw)(T2-T3)\n\nwhere c1 is the specific heat capacity of the metal, and cw is the specific heat capacity of water (look it up on Google to get the units you need).\n\nSolving this, you get:\n\n......(M2)(cw)(T2-T3)\nc1 = -----------------\n........(M1)(T3-T1)\n\n(sorry about the dots, I'm trying to keep everything aligned)\n\nRepeat the calculations for each of the metals.\n\nYou will need to calculate (or look up) the densities of each of the metals.  Then, graph density vs. the specific heat capacities you calculated, and see if your hypothesis is true.\n\nHave fun!
408671
its not like that , but according to HVAC data , the temp. underground is constant about 10 degrees celsius , so in summer its hot , then we find the water cooler , and in winter , the air temp. may be zero , so the water is hotter
339053
I've ran a pool for 2 summers now.  People seem to always say that 81-82 feels cool, 83 is warm (just right) and 84+ feels too hot, but we do have a lot of little kids, and it's in Seattle.  But those temps hold true pretty much whatever the air temperature is.
113745
You should be able to find the answers on these web pages:\n\nhttp://www.uni.edu/~iowawet/H2OProperties.html\nhttp://dl.clackamas.cc.or.us/ch106-06/thermal.htm\nhttp://daphne.palomar.edu/jthorngren/latent.htm
196969
its 0 degrees c\nand 32 degree F\n\nBut this is assuming it normal tap water without to much sediment like rust.  I don't know what temp sea water freezes at.
912037
Find a question that relates to the local area you live in, something you may have noticed, or has controversy.  A saying I like is "One measurement is worth a thousand expert opinions", which is the essence of science.  So find something to measure.  When I was in school (a long time ago) people I knew wondered about whether the local creek was clean enough.  Kids played in the creek and some people fished in it, but I thought it shouldn't foam and have a not quite clear look.  So, I got some sterile petri dishes from the biology lab, got some q-tips, and talked the local hospital pathologist into helping me.  If you need help, ask, you might be surprised, it turned out he was curious too.  Anyway, I learned a bit about how to identify bacteria and yes, the creek was polluted-bad!  I made the front page of the local paper.  I'm not saying do this, but think of a local problem and measure it.  Hope this helps.