[bondxazn]

the vapor pressure above a pure liquid or solid is
given by the Clausius-Clapeyron equation,
where for liquid to vapor we use the molar enthalpy of vaporization, and for solid to
vapor we use the molar enthalpy of sublimation. The molar enthalpy of vaporization for
water is ΔHvap,m = 40.7 kJ/mol at 100 °C and 44.0 kJ/mol at 25 °C.
A 2.00 kg block of solid copper (Cp,s = 0.385 JK–1g–1) is removed from an ice bath,
quickly dried, and introduced to a large insulated dewar of volume V containing 1.00 mol
of water vapor at 100 °C an 1 atm pressure (I guess with gloves(!) as the steam would
burn you to a crisp!).
(a) (1 pt each) Assume all the steam is condensed into water (this is untrue,
as there is always some vapor, but it is close enough). What will be (i) the final
equilibrium temperature; (ii) the heat transferred from the water to the copper; (iii) the
entropy change of the water; (iv) the entropy change of the copper; and (v) the entropy
change of the whole system for this spontaneous process? (You may assume the specific
heat of water is 4.184 JK–1g–1 at all temperatures.)
(b) (1 pt each) In fact, some water vapor is present at equilibrium, because
water has a nonzero vapor pressure. Assuming that steam behaves like an ideal gas, that
2 kg of Cu has negligible volume, and that the molar heat capacity of water vapor is Cp,m
= 37.3 JK–1mol–1 calculate (i) the volume V of the original dewar (see above); (ii) the
amount of water vapor that will be present at the final temperature found in part (a); (iii)
the fraction of the water that should be present as vapor at equilibrium.
(c) (2 pts) Will the final temperature of the real system be higher or lower
than what you figured in part (a)? Carefully justify your argument.

[MOTOBALL]

The principle for all of these "add object A at Ta deg. C to object B at Tb deg C and what is final eqlbm. temp Teq ?" is

Heat gained by A = heat lost by B

So, assume here that A is at O C deg (Cu block) and B is at 100 C (steam)

Heat gained by A is (Teq  - O ) x specific heat x mass

Heat lost by B is [(100 - Teq) x latent heat of vapzn. of water x mass] to condense the steam PLUS [(100 - Teq) x specific heat of water x mass] to cool the water.

This is the first thing to solve, and is why it is given as problem (i) in part (a).  Everything else can then be determined.