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

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.

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

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`

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

To calculate the amount of heat required to convert all the liquid water into vapor at a higher temperature, we need to use the heat transfer equation:

q = m * ΔH

where:
q is the heat required (in J),
m is the mass of the substance (in grams), and
ΔH is the molar enthalpy of vaporization (in J/g).

First, we need to determine the mass of water (m). Given that we start with 146 g of water, this will be our value for m.

Next, we need to find the molar enthalpy of vaporization (ΔH). The molar enthalpy of vaporization of water is 40.7 kJ/mol. To convert this to J/g, we need to divide by the molar mass of water, which is approximately 18 g/mol.

ΔH (in J/g) = (40.7 kJ/mol) / (18 g/mol) = 2.26 kJ/g

Now we have all the necessary information to calculate the heat required (q):

q = m * ΔH
q = 146 g * 2.26 kJ/g
q ≈ 331 kJ

Therefore, you would need to add approximately 331 kJ of heat to convert all the liquid water into vapor at 100◦C.

Please note that this calculation assumes no heat loss to the surroundings, which may not be entirely realistic in practical scenarios.