31.47mol of copper at 273 kelvin put inside an isolated cup along with 1 mol of water vapors at 373 Kelvin.
(pressure is constant at 1 atm).
ALL of the water condensed.
Cp(Cu(solid)) = 24.44 J/mol
Cp(H20(gas) = 33.58 J/mol
Cp(H20(liquid) = 73.35 J/mol
dHcondensation of water = -40,700J/mol.
a) find the final temperature of the system
b) what is the amount of heat transferred from copper to the water
c) find entropy change for water, copper and the whole system.
The Attempt at a Solution
here is what im thinking.
This is isolated system so no heat is coming out.
the heat given from the water = the heat given from the condensation which is -40700J plus the heat given from cooling the water which is Q = Mass(water)*Cp(liquid)*(T(final) - 373K) = 75.35J/K (Tf - 373) Tf = final temperature.
Now the heat given from the copper (should be negative since the copper receive the heat from the water) is Q=Mass(copper)Cp(copper)*(Tf - 273) .
now since its isolated, the heat from the copper should be equal to the heat from the water but when summing it all up, i get that the final temperature is less than the starting temperature of the copper which is not possible ofcours.
thank you so much in advance.