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`