Question

How much heat is required to warm 1.40 l of water from 20.0 ∘c to 100.0 ∘c? (assume a density of 1.0g/ml for the water.)?

Answer 1

The water density is 
$d=1.0 g/mL = 1000 g/L$
And the mass of 1.40 L of water is
$m=dV=(1000 g/L)(1.40 L)=1400 g$

The amount of heat needed to increase the temperature of the water by $\Delta T$ is given by
$Q=m C_s \Delta T$
where m is the water mass, $C_s = 4.18 J/g ^{\circ}C$ is the water specific heat capacity and 
$\Delta T=100.0 ^{\circ}C-20.0 ^{\circ}C = 80.0^{\circ}C$ 
is the increase in temperature. If we substitute these numbers into the equation, we find
$Q=(1400 g)(4.18 J/g^{\circ}C)(80.0^{\circ}C)=4.68 \cdot 10^5 J$