Is the density of ice water equal to the density of ice or is it still a little more? My book give s density of ice = 917 kg / m^3 and water at 4 degrees at 1000 kg / m^3. for ice water do i still use the 1000?
202386
Water is water. The density of ice water is 1000 kg/m^3. It isn't floating is it? Then it must not be below 1000.
552534
Fluid Pressure= weight density x (times) depth\n\nFl. p.= 10,000 kg/m^3(thats the weight density of water) x 15 m\n\nFl. p.= 150,000 m^2 (because one m cancels)\nHope this helps!
543539
Looks like this is your homework. anyway, i'll only answer two of your questions.\n\n1. the density of water is almost 1 g/mL. and 1 liter = 1000 mL. hence the mass of 1 liter of water is 1000 g.\n\n2. since 1 liter of water is 1000 g or 1.00 kg, its weight is 1.00 * 9.8 = 9.80 N
164860
Salt, NaCl, turns into Na+ and Cl- ions in water.\n\nThe ions changes the equilbrium of H2O melting and freezing.\nLiquid water is block by the ions, so it has a harder time freezing back onto the ice. Since the ions doenst slow down\nthe melting of ice, but slows down the freeze of water.\nThe ice now melts at below 0 degree.\n\nSo a normal ice water will be at zero degree.\nBut a salt put on the ice/water will melt more of the ice at a lower temp., about 7-10 degree colder, depenting on the ammount of salt and ice avalibale.
663664
It's been awhile since i've done this but i'll give it my best\n\nFirst off, the first energy you need to calculate is the energy needed to raise the temperature of ice to 0 degrees C. so...\n\n(12.37 g)(.48 cal/deg/g)(35 deg) = +207.816 cal\n\nWhen the ice changes to water you have a positive heat of fusion. So...\n\n(12.37 g)(80 cal/g) = 989.6 cal\n\nNext you have to raise the water temperature from 0 degrees C to 100 degrees C. So...\n\n(12.37 g)(1.0 cal/deg/g)(100 deg) = 1237 cal\n\nNext you have the positive heat of vaporization.\n\n(12.37 g)(540 cal/g) = 6679.9 cal\n\nNext once it is steam, you must raise the temperature from 100 degrees C to 146 degrees C. So...\n\n(12.37g)(.48 cal/deg/g)(46 deg) = 273.1296 cal.\n\nNow that you have the energy for each step add them together.\n\n207.816 + 989.6 + 1237 + 6679.9 + 273.1296 = 9387.45 cal\n\nI'm pretty confident in my breakdown but it defintely wouldn't hurt to have someone else check it
193800
It is usual for liquids to contract (about 10%) on freezing and expand on melting. This is because the molecules are in fixed positions within the solid but require more space to move around within the liquid.\n\n\nWhen water freezes at 0°C, at atmospheric pressure, its volume increases by about 9%. If the melting point is lowered by increased pressure, the increase in volume on freezing is even greater (e.g. 13% at -20°C). The structure of ice  is open with a low packing efficiency where all the water molecules are involved in four straight tetrahedrally-oriented hydrogen bonds;
84465
Specific gravity.\n\nPure water has a density of 1 (the mass of 1 cubic centimetre of pure water is precisely 1 gramme), if something was twice as heavy as water (i.e. it has twice the mass) it would have a specific gravity of 2, 6 times as heavy and the SG would be 6.  If something was half as light as water it's SG would be 0.5 and so on.
444766
It is just one of the properties of the H bonding of water and the type of crystal lattice it forms. Also not all ice is less dense than water.  If you froze water with tritium or deuterium instead of regular H the ice would sink.\n\nAlso water reaches it maximum density as it approaches its freezing point.
284165
D) It's heat of fusion is higher than its heat of vaporization.\n\nWater's heat of fusion is: 334.5 kJ/kg\nWater's heat of vaporization is: 2256 kJ/kg\n\n\nE IS NOT CORRECT BECAUSE: Water is denser than ice. If water was not denser than ice, ice would not float.\n\nAnd Mr. the_bendude, you must have never taken chemistry. Heat is not the same as temperature. Yes, water boils at 100C and freezes at 0C, but that does not mean those are the heats for which it does.
106422
Relative density (also known as specific gravity) is a measure of the density of a material. It is dimensionless, equal to the density of the material divided by the density of water (or, sometimes used for gases, of air).\n\nSince water's density is 1.0 × 103 kg/m3 in SI units, the relative density of a material is approximately the density of the material measured in kg / m3 divided by 1000 (the density of water). There are no units of measurement.\n\nWater's density can also be measured as nearly one gram per cubic centimetre (at maximum density) in metric units. The relative density therefore has nearly the same value as density of the material expressed in grams per cubic centimetre, but without any units of measurement.\n\nRelative density or specific gravity are often ambiguous terms. This quantity is often stated for a certain temperature. Sometimes when this is done, it is a comparison of the density of the commodity being measured at that temperature, with the density of water at the same temperature. But they are also often compared to water at a different temperature.\n\nRelative density is often expressed in forms similar to this:\n\nrelative density:  or specific gravity:  \nThe superscripts indicate the temperature at which the density of the material is measured, and the subscripts indicate the temperature of the water to which it is compared.\n\nDensity of water calculated from formula in 68th CRC Handbook of Chemistry and Physics 1987–1988.\n\nDensity of water at 1 atm (101.325 kPa, 14.7 psi)\n(little or no difference at 100 kPa) Temperature Density \nCelsius Fahrenheit kg/m³ \n0 °C 32 °F 999.83952 \n3.98 °C 39.16 °F 999.9720 \n15 °C 59 °F 999.0996 \n16 2/3 °C 62 °F 998.8322 \n20 °C 68 °F 998.2041 \n25 °C 77 °F 997.0449 \n \n\nWater is nearly incompressible. But it does compress a little; it takes pressures over about 400 kPa or 4 atmospheres before water can reach a density of 1000.000 kg/m³ at any temperature.\n\nRelative density is often used by geologists and mineralogists to help determine the mineral content of a rock or other sample. Gemologists use it as an aid in the identification of gemstones. The reason that relative density is measured in terms of the density of water is because that that is the easiest way to measure it in the field. Basically, density is defined as the mass of a sample divided by its volume. With an irregularly shaped rock, the volume can be very difficult to accurately measure. The most accurate way is to put it in a water-filled graduated cylinder and see how much water it displaces. It is also possible to simply suspend the sample from a spring scale and weigh it under water. Solving Isaac Newton's equations yields the following formula for measuring specific gravity:\n\n \nwhere G is the relative density, W is the weight of the sample (measured in pounds-force, newtons, or some other unit of force), and F is the force, measured in the same units, while the sample was submerged. Note that with this technique it is difficult to measure relative densities less than one, because in order to do so, the sign of F must change, requiring the measurement of the downward force needed to keep the sample underwater.\n\nAnother practical method uses three measurements. The mineral sample is weighed dry. Then a container filled to the brim with water is weighed, and weighed again with the sample immersed, after the displaced water has overflowed and been removed. Subtracting the last reading from the sum of the first two readings gives the weight of the displaced water. The relative density result is the dry sample weight divided by that of the displaced water. This method works with scales that can't easily accommodate a suspended sample, and also allows for measurement of samples that are less dense than water.
512367
If we consider the volume of water is known.(If not known then use graduated cylinder and find the volume)\nMass of water can be calculated using following formula:\n\nDensity(ρ)=mass(m)/volume(V) , where\nρ is the object's density (measured in kilograms per cubic meter) \nm is the object's total mass (measured in kilograms) \nV is the object's total volume (measured in cubic meters) \n(Temp (°C)/ Density (g/cm3) \n30/ 0.9957 \n20 /0.9982 \n10 /0.9997 \n0 /0.9998 \n−10/ 0.9982 \n−20 /0.9935 \n−30/ 0.9839 )
290972
Most liquids dilate when heated and therfore the density increases with temperature. Water is an exception below 4 degrees Celsius. The reason are hydrogen bonds. \n\nIn the H20 molecule the hydrogen atoms share their electrons with the oxygen atom in socalled covalent bonds. But the oxygen atom attracts the electrons very strongly and takes more than its share.\n\nAs a consequence the hydrogen atoms form bonds with the oxygen atoms of other water molecules. These hydrogen bonds are weaker than the covalent bonds. \n\nThe result of the existence of hydrogen bonds is that ice is organised in a hexagonal structure which is less dense than water. These structures form already in liquid water too, but to a lesser degree.  \n\nThe end result is a tug-of-war between the effect of the hydrogen bonds and the normal dilation of water. With a maximum density at 4 degree C as a consequence.
20713
1) We need to raise the temperature of the ice from -16.73°C to 0°C, which is a temperature rise of 16.73°C.\n\nMass of ice * specific heat of ice * temperature rise\n38.80g * 2.1 J/g K * 16.73° = 1,292.89 J\n\n\n2) We now need to add heat until all the ice is melted:\n\nMass of ice * Heat of Fusion\n36.80g * 333J/g = 12,254.40 J\n\n\n3) Now that the ice is all melted, it's all water, but the temperature is still 0°C. We now have to raise the temperature of the water to 100°C, which is a temperature rise of 100°.\n\nMass of water * specific heat of water * temperature rise\n36.80g * 4.2 J/g K * 100° = 15,456.00 J\n\n\n4) Now we have to change all the water into steam.\n\nMass of water * Heat of Vaporization\n36.80g * 2260J/g = 83,168.00 J\n\n\n5) All the water is now steam, but the temperature of the steam is still 100°C. We now have to raise the temperature of the steam to 227.6°C, which is a temperature rise of 127.6°.\n\nMass of steam * specific heat of steam * temperature rise\n36.80g * 2.0 J/g K * 127.6° = 9,391.36 J\n\n\n\nNow add up all the Joules\n1,292.89 J + 12,254.40 J + 15,456.00 J + 83,168.00 J\n + 9,391.36 J = 121,562.65 J\n\nConvert to kJ = 121.56kJ
403278
Density is a measure of mass per volume. Same amount of water; sugar water is denser. Weigh; salt is NaCl, sugar is a more complex and heavier molecule. Formula for glucose. C6 H12 O6. Denser.\n\nNo. sugar water is denser than salt water. Period.
554379
The formula for density is mass/volume.\n\nFind the density of each liquid using the formula above:\n\nDensity of liquid A = 48 / 30 = 1.6 g/mL\nDensity of liquid B = 20 / 20 = 1 g/mL\nDensity of liquid C = 8 / 10 = 0.8 g/mL\n\nSo to put them in order of increasing density:\n\nC < B < A\n\nI hope this helps!
40041
>>>>>at 4 degrees above its freezing point, water sinks.\n\nThis is true.\n\n>>>>any lower, and it begins to freeze, forms crystals, and floats.\n\nThis is not true. Water begins to freeze at 0oC, not at 3.9oC or even 1oC.\n\n>>>>any warmer, and the molucular action makes it less >>>>dense.\n\nThis is not true. Water becomes less dense at warmer temperatures due to kinetic motion impeding the formation of hydrogen BETWEEN molecules.
312200
density = mass/volume.\n it changes wt respect to temp.\n as temp increases,volume increases and so density decreases. however there may b some exceptions.\n  as an example, if v heat water to high temp but less than it's boiling point, it's volume increases.\n  so, v can say that the density is less than the previous condition.