Question

How much heat is required to convert 2.55g of water at 28 degrees c to steam?

Answer 1

There are two different processes here:
1) we must add heat in order to bring the temperature of the water from $28^{\circ}C$ to $100^{\circ}C$ (the temperature at which the water evaporates)
2) other heat must be added to make the water evaporates

1) The heat needed for process 1) is
$Q_1=m C_s \Delta T$
where 
$m=2.55 g$ is the water mass
$C_s = 4.18 g/J^{\circ}C$ is the water specific heat
$\Delta T=100^{\circ}C-28^{\circ}C=72^{\circ}C$ is the variation of temperature of the water
If we plug the numbers into the equation, we find
$Q_1 = (2.55 g)(4.18 J/g^{\circ}C)(72^{\circ}C)=767.4 J$

2) The heat needed for process 2) is
$Q=m L_e$
where 
$m=2.55 g$ is the water mass
$L_e = 2264.7 J/g$ is the latent heat of evaporation of water
If we plug the numbers into the equation, we find
$Q_2=(2.55 g)(2264.7 J/g)=5775.0 J$

So, the total heat needed for the whole process is
$Q=Q_1+Q_2=767.4 J+5775.0 J=6542.4 J$