Insulated Travel Physical Review 10 some questions?
5cm side and the aluminum cube 1.An º 500 is the temperature of the heat. After it is placed on travel Well insulated (no heat a cup of water cup of 500cm3) is put into. º mulbuteo To 10 C 10000L tanks heating 5000W of heat received from the search, it Will. 2.How long the water, the final temperature of the cup 30 º? The other side 3.A At room temperature (23 º C while the radius of 5cm and length of 2m and copper rods An end to 425 º C) is heated. How much heat from the hot end of 100J transport will take you down at the end of the bar? 4.An aluminum rod and C º 2 minutes long iron bar in an ice bath (0) If water is taken With all the heat out of the boiling 100 C º, what is the difference in length? 5.A of steel tanks with capacity of the C º 75L 94 will be filled ehaneun If it is at the top with hot water, water 4 º C is much older If the water tank overflow? Can anyone me, thank you It helps wih
1. We are the one .. If you want to re-post Questions can not work out = Jean ==> but you Did not specify the initial temperature of the water as long as 2 2. Energy / Hour, so t = power = E / P, where E = m * C * deltaT the = 10000L * 1kg/1L * 4200J/kg-oC Number * (30 -10) 8J, so t = ^ a = 8.4x10 8.4x10 ^ 8 / 5000 = 1.68x10 ^ 기들 = 46.7hr 3. H a = Q / t = k * (beautiful - TC and) / L = 385W/mK * pi * (0.05m) ^ 2 * (425-23) / 2 = 608W, so that t = Q/608 = 100J/608W = 0.165s 4. Of alpha * L0 * deltaT --- DeltaL (Al) - deltaL (Fe) = difference Length = DeltaL = 2m * 100 * (alpha-Al - alpha iron) = 200 * (2.4x10 ^ -5 - 1.2x10 ^ -5) = 0.0024m = 2.4mm 5 using Beta version of the DV = V0 * * Products DT = 210x10 ^ -6 * 75L * (94-4) = 1.42L
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