Question

How much heat is required to warm 1.30 l of water from 24.0 ∘c to 100.0∘c? (assume a density of 1.0g/ml for the water.)?

Answer 1

When q = M* C *ΔT 

we can get the mass using the given volume & density:

M = volume * density 

   = 1.3 L * 1g/ml * 1000mL1L  

   = 1300 g 
and when the  specific heat of water = 4.18J/g.°C

now, by substitution:

∴ q = 1300g * 4.18J/g.°C * (100-24)

   = 412984 J or 412.984 KJ