What are the utilities of a ice-pack? (all the utilities)?
609524
You can make your own ice pack by combining one part rubbing alcohol and two parts water, then filling a plastic bag with this (two, preferably) and freezing it. The alcohol keeps the pack from freezing completely, so you can mold it to any part of your body.
944831
the first thing to do is figure out how much heat exchange it takes to get the water to 0C.\n\n12kg x 10 = 120 Calories\n\nnow figure out what 120 Calories will do to the ice\n\nit takes 8.1 x 5 = 40.5 Calories to get the ice to 0C\n\n120 - 40.5 = 79.5 Calories are available to melt the ice\n\nheat of fusion for water is 80 Calories/kg\n\n79.5 / 80 = 0.99375 kg ice melts\n\n8.1 - 0.99375 = 7.1 kg ice\n12.0 + 0.99375 = 13.0 kg water
825790
Elemental Estimating and Cost Planning\n\nThe system allows for input of elemental or unit price information from previous projects. \n\nThese may be broad detailed elemental costs and may be from estimates, tenders or final accounts.\n\nThis information can also be automatically transferred from ICEMATE's sister product ICEPAC.\n\nThis indexed cost information can then be accessed at various levels, selected and manipulated to create new estimates and cost plans. Analytical reports and formal documentation can then be produced.\n\nElemental costs are expressed by user-defined parameters at each of various levels. For example, the project may be expressed in cost per kilometre or per head of population; each part may be expressed per metre or megalitre/day. Similarly each sub-part, group or element/sub-element may have differing yardsticks by which to express cost.
577683
http://en.wikipedia.org/wiki/Ice-T\n\nTry this it should help!!!
67218
Water to drink\nWood for heat\nWeather (shelter from it)\nWigglies (no snakes or creatures) \nWidowmakers (nothing big that could fall on you)
938830
take some water and put it in the freezer, go and play some video games for 4 hours, come back, and you will have made ice
272305
An ice house..., covered in saw dust lasts all summer..., same with snow..., tree planters cover their trees in winter with snow then saw dust..., snow will last right til August / September...., so just put your cube in there...\n\nthere are many poor conductors for heat..., what I would do is wrap the cube in tin foil..., then wrap in white plastic (some people refer this as a cooler) or white foam..., etc....
717469
I dont understand your question.  How can things work without electricity?  Electricity is power.  And power is essential for any instrument to work.  Do you mean to ask about using alternative electricity?\n\nYeah you can put the fridge in snowpacked mountains which will make things inside cool.
803746
Cool,\n-make it arrange 3 boxes on top of each other\n-make it find the power socket\n-trace and stand beside tabl where your glass (attached sensor) located\n-shut the door for you\n-find your shoose maybe clean it\n-play with your pet\n-get you bear can from fridg ;)\nlet me know what you did it's so interesting subject for me.\nlabibbaroudi@hotmail.com
730140
The ice would gain 124.3 kJ in moving from -10 °C to zero.\nThe water would lose 252.0 kJ in moving from +10°C to zero.The difference in the two energies is sufficient to melt 383 g of ice.\nSo the mixture will be at 0 °C and will consist of 6383 g of water and 5617 g of ice.
369633
try http://packing.org
419872
Anything Coleman. There coolers rock!!!!
838527
no need of ice cubes \nto pu in trousers\njust sit in refrigerator for 5 min
23724
Yes.  Start > All Programs > Accessories > System tools  and there you will find the various utilities.
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
135368
A thermos.\n\nAn actual thermos (or dewar flask) has a vacuum seal that allows no heat to transfer through (because it's empty), thus minimizing the heat loss or gain.
567556
Well, 1st thing, how much energy you want to put into it?