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?

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