Chemistry, Vaporization Help!!!?

Question

A 50.0g piece of iron @ 152 C is dropped into 20.0g H2O(l) at 89 C in an open, thermally insulated container. How much water would you expect to vaporize, assuming no water splashes out? The specific heats of iron and water are 0.45 and 4.21 J/ g C , respectively, and Hvaporization = 40.7 KJ/mol H2O.

Help!!!


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Answer 1

First of all, assume constant pressure of 1.0 atm. This is mostly importantfind, because temperature of vaporization depends on it. Accordingly, assume vaporization occurs at the constant temperature of 100 C and at the end of the process, all bodies and phases will be at that temperature. So, first the water is at 89 C and heats up to 100 C where it starts boiling, if pressure is constant, boiling occurs at that constat temperature. Now, calculate the heat (Call it QW) needed to take water from 89 to 100 C using:

Q = m Cp (T2 - T1)

´Q=amount of heat
m is the mass
(Cp is specific heat)
T2 is the final temperature of the process
T1 is the initial temp of the process

This heat is commonly called sensible heat, meaning that it manifest as a change of temperature

Now, using the same expression calculate the amount of heat the iron loses as it cools from 152 to 100 C (Call it QFe) So this result must be greater in magnitude than QWcalculated before, but it will be negative, meaning that the iron body is loosing energy.

Now the heat of vaporization of water would be;

Qv = n Hv

where n = moles of water vaporized
Hv = you know what it is

now

n = Qv/Hv

Make the energy balance: The heat lost by the iron body when cooling, is equal to the heat gained by water Qw to increase its temperature and the heat to vaporize a certain amount of water

QFe = Qw+Qv

So:

Qv = QFe - Qw (All heats must be positive at this point)


obtain the moles of water n by using the expression given before

If you want, you can obtain the mass of water multiplying moles by molecular weight