If you start with 146 g of water at 47◦C, how

Question

much heat must you add to convert all the
liquid into vapor at 100◦C? Assume no heat
is lost to the surroundings.
Answer in units of kJ.

Answer 1

q1 = heat needed to raise T from 47 C to 100 C.
q1 = mass H2O x (Tfinal-Tinitial)

q2 = heat need to vaporize H2O at 100.
q2 = mass H2O x heat vaporization.

Total = q1 + q2

Answer 2

We have to look first for the specific heat capacity and latent heat of vaporization for water. These values can be found in textbooks or in google.
Specific heat of water, c = 4.18 J/g-K
Latent heat of vaporization, Hv = 2256 J/g

The first process is heating of the water from 47 deg C to 100 deg C (which is the boiling point of water). This heat is called the sensible heat. It just heats the water, but it does not change its phase. We use the formula:
Q = mc(T2-T1)
where
m = mass (g)
c = specific heat (J/g-K)
T = temperature (K)
Substituting,
Q = 146 * 4.18 * (100 - 47)
Q = 32344.84 J

The second process is the changing of phase of water from liquid to vapor. This heat is called the latent heat. Note that this is a temperature-constant process. We use the formula:
Q = m*Hv
where Hv = latent heat of vaporization
Substituting,
Q = 146 * 2256
Q = 329376 J

Finally, we add the two Q's:
32344.84 + 329376 = 361720.84 J

Just convert to kJ as the problem asks.
hope this helps~ `u`

Answer 3

q1 = mass H2O x specific heat H2O x (Tfinal-Tintial)