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

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 

The answer is d.potassium