How much heat is required to warm 1.20 l of water from 26.0 ∘c to 100.0 ∘c? (assume a density of 1.0g/ml for the water.) ?

When we have the volume = 1.2 L

and density = 1000 g/L

∴ mass of water (m) = volume * density

= 1.2 * 1000

= 1200 g

according to this formula, we can get the amount of Heat q:

q = m * C * ΔT

= 1200 g * 4.18 * (100 - 26)

= 371184 J