-40oF =[ (-40)-32}/1.8]oC = -40oC
Ice had latent heat, it could give more of chilling effect than putting the same quantity of water at 0 oC. For example, a 330 cc of cola at 25 oC, drop in 100 g of ice and you got it to about 1 oC. Blend in with icy water and you got it at 19 oC. Blend in icy water to 1 oC and you got yourselves bucket of water with faint taste of cola. Have some calculation, water heat capacity is 4.2 kJ/kg.oC and heat of fusion is 333 kJ/kg.
Water is a liquid between 0 oC and 100 oC.
q = m x C x ΔT q = amount of heat energy gained or lost in calories m = mass of substance (in this case water) in grams = 10g C = heat capacity of substance (in this case water) = 1cal/gram•oC Tf = final temperature = 32 oC Ti = initial temperature = 0 oC ΔT = (Tf - Ti) = 32 oC q = 10g x 1cal/gram•oC x 32 oC = 320 calories
Many instruments and volumetric glassware are calibrated at 20 oC.
20 oC.
1oF change = 5/9 oC change 13oF = 13 x 5/9 oC = 65/9 oC = 7 2/9 oC ~= 7.22 oC
141 degrees is an obtuse angle. 141 oC = 285.8 oF and 414.15 K. 141 oF = 60.555... oC and 333.70555... K. 141 K = -132.15 oC and -205.87 oF.
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27 oF = -2.8 oC = 270.4 kelvin.
oC = 5/9 (oF -32)oC = 5/9 (122 - 32)oC = 5/9 ( 90 )oC = 50
-40oF =[ (-40)-32}/1.8]oC = -40oC
259 K equal -14 oC.
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-42 oF = -41.1 oC
A full port OC48 has a capacity measured in GB. The capacity is 2.884 Gbps for OC48. Converted into Mbps this would be 2,547.712 Mbps.
276.7872 j I s the answer