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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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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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