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29 March, 13:02

A 6,000 liter tank of water is at 0 oC. It must be heated to 20 oC. How many calories are required to do so?

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  1. 29 March, 14:01
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    The number of calories that are required to do so is 1.2 x10^8 cal

    calculation

    Heat = MCΔT

    where M=mass in grams (gotten by converting L into g)

    where 1 L = 1000 g

    6000L=? g

    by cross multiplication

    = [6000 x1000] = 6,000,000 g

    C = specific heat capacity for water = 1 cal/g/c ΔT=change in temperature = 20c-0c = 20 c

    therefore heat = 6,000,000 g x 1 cal/g/c x 20 c = 1.2 x10^8 cal
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