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A sample of water in a calorimeter has a mass of m = 0.22 kg and is at a temperature of Ti = 21.4 C. The voltage across the heating coil is V = 7.1 V and the current through it is I = 8.5 A. Assuming that no heat is lost to the environment or the calorimeter itself, how long will it take (in hours) to boil all of the water into steam? The latent heat of vaporization for water is Lv = 2.26e6 J/kg.
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Added Tue, 04 Aug '15
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Heat Power generated across coil = VI = 7.1 X 8.5 = 60.35 W

Enegry required to raise the water temperature from 21.4C to 100C

= mass X specific heat capacity of water X Temp Diff

= 0.22 X 4181 X (100 -21.4) = 72.3 KJ

Enegry required to vapourise the water = mass X latent heat of vaporization

= 0.22 X 2.26 e6 = 4.97 e5 J

Total energy required =  72.3 KJ + 4.97 e5 J = 570 KJ

Time = Total energy required / Heat Power generated

= (570 X 1000) / 60.35 = 9436.62 s = 2.62 hrs
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