Calculate the temperature of the water at the bottom of the falls?
919581
The potential energy mgh of 1 gram of water 0.001 x 9.813 x 50 = 0.49 Joule. \nThe specific heat of water is 4.2 J/g/K, therefore 0.49 Joule raises the temperature of 1 gram of water 0.49/4.2 = 0.12 degree.\n\nThe  temperature of the water at the bottom of the falls is 10.12 C
554775
Put it in a graduated cylinder, record the water level, drop the rock in, and note the new water level. Subtract.\n\nYay science.
1165990
http://www.temperatures.com/apps/index.html
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.
811304
let thetemperature of the mix be 't'*Cfind the heat lost by coffee\nfind the heat gained by ice in three stages\na)from -2.2*C to 0*C\nb)for melting and becoming water at 0*C\nc)for getting to the equilibrium temperature of the mixture\nequate and solve
160584
Um...well first you take some water, add in a cliff of some sort, then lastly you sprinkle on some gravity.  Voila!  You have yourself your very own waterfall.  It has water, and guess what?  It's falling!
328342
Simply d1/ d2 = weight of metal / loss of weight in water = 10/ 2 = 5.\n\nd1 = 5x10^3 Kg/ m^3\n  \n\nDetails\n\nThe reduction in downward force is 2N.\n\nThis is due to the upward force given by equal volume of water. \n\nIf V is the volume and its mass is V.  10^3 ( density of water is 10^3 kg/m^3).\n\nThe weight of this volume of water is V.  10^3 g This is equal to 2N. \n\nThe weight of the metal is V. d . g ( d is the density of the metal) = 10N.\n\nTherefore,  V.  d . g/ V.  10^3 g ==10/ 2 = 5\n\nd = 5 x10^3 kg/m^3.
911974
Go with a line graph, that way you can also find out (or estimate) temperatures at times when you didnt measure it.
658781
Freezing and Boiling points of Water are:\n\n 0 and 100 degrees Celsius\n32 and 212 degrees Fahrenheit\n\nFrom the equation of a straight line,\n\nF = mC + A\n\nwhere m is the gradient and A is a constant\n\nm = (212-32)/(100-0) = 1.8\n\nF = 1.8C + A\n\nwhen x=0, F=32\n\n32 = 1.8(0) + A\n\nA = 32\n\nTherefore the final equation is:\n\nF = 1.8C + 32\n\nor\n\nC = (F - 32)/1.8
850698
The fall time T = sqrt(2*h/g) = 2.22 s. So only 4.22-2.22 = 2 s of water poured is in the pail. That's 2*.111 kg added to 1.07 kg = 1.292 kg. That mass * g = 12.66 N.
108765
hi there,\nits like, you need to know the temp and the mass flow rate of the bottom product of the distillation column. also you need to know the boiling point of the bottom product. the mass flow rate depends on your design calculation. temp of the bottom product you should already know. and boiling point of the bottom product is readily availble.\n\nnow use m.cp.dt to find heat required to raise the temp of the product to its boiling point. where cp is the specific heat capacity of the bottom product and is available in handbooks or sometimes expressed as functions of temperature itself. and dt stands for differnce of boiling point of bottom product and the current temp of the bottom product.call this value as h1 (say).\n\nnow you know the latent heat of vapourisation of the bottom product again refer to data handbooks for the values. multiply this by the mass flow rate. and this is the heat required to vapourise the bottoms. lets call this h2 (say)\n\nnow h1+h2, gives you the heat load for the reboiler. use this heat load to calculate the steam requirement etc...\nyou can mail me sanjaykumar_16@yahoo.com for any clarifications.
580754
i haven't done this in a while.\n\nThe volume of the reservoir is equal to (1/2)*(height)*("base")*length.  the reservoir is a rectangular prism with the "base" being the distance across the surface of the water.  We are given the shape of the full volume so we can scale down to the level of the water to find what the volume in the reservoir at its current level.\n\nif you think about the cross-section of the volume, and use half of that forming a right triangle, we can determine the ratio of the surface (the "base") to the height (water depth).  In this case, it is (40/2)/(20*sqrt(3)) or 1/sqrt(3).  so ("base"/2)/(height) = 1/sqrt(3)  and then "base" = (2*height)/sqrt(3).\n\nsubstitute that into the volume equation so that\nvolume = [(height^2)*length]/sqrt(3) or literally,\nV = [(h^2)*L]/sqrt(3).  differentiate with respect to time,\n(dV/dt) = [2*h*(dh/dt)*L]/sqrt(3) and substitute the numbers in:\ndV/dt = 200 - 50 = 150\nh = 10*sqrt(3)\nL = 450\nand solve for dh/dt which is the rate of change of the water level in yds/hour.  these are odd units, you might want to change to inches/hour.\n\ni did it this way and got 1/60 yds/hr or 0.6 inches/hr.\n\nbut again it has been a long time....
222680
temp 30-34 deg C\nadequate Nitrogen & Phosphorus\nwell watered but well drained\nNot too Humid\nC02 level 1000ppm
182862
First, weigh the liquid or solid. Then measure the temperature and record both values. Now, you need to set up a current. You need ampermeter and voltmeter in the current.(Make sure the current is not shorting, i.e. touching the ground.) Then run the current through the solid or liquid and measure the temperature after you are done. Calculate the electrical energy that you gave to the liquid or solid. \n\nFind heat capacity: Energy / change in temperature.\n\nThen find specific heat capacity: heat capacity / mass.\n\nIf you can heat it up, that way easier than an electrical method. If you have to use an electrical method, you need a lot of calculations, measurements, etc.\n\nGood luck!!!
169646
The water pressure t a given depth is give as the weight of the water above that depth divided by the area it is being applied.\n\nSo imagine a column of water extending from the top of the lake down the depth.  Calculate the volume of this column of water then using the density of water (1 grams / mL) find the mass.  The take the mass of the water and find its weight (W = mg).  The weight it is the force exerted by the water, but pressure is the force of the water divided by the area it is applied.  So divide the force by the area of the base.\n\nEDIT:\nI see you added the density of water (even though I think my density [1 g/cc] is better).  I will leave you to do your own homework.\n[I also refuse to work with such an ugly unit, I prefer metric]
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.
505556
you have to look for it at the saturated watter boards or the saturated water graph, then the temperature will tbe the saturation temperature, according to my graph, the temperature would be...\n\n83.79ºC