A block of copper of mass 2.00 kg and specific heat capacity Cp=0.385 J/K·g at a temperature of 0°C is introduced in an insulated container in which there is 1 mole of water vapor at 100°C and 1 atm. What is the final temperature of the system at equilibrium assuming no heat transfer to the surrounding?

Question

anonymous · Answer

first you need to notice that same heat that is released by water is the heat gained by copper and now on it is pretty simple...

Q = m cp dT

where index 1 is the index of copper and index 2 is index of water

so Q1=Q2

m1 cp1 dT = m2 cp2 dT + dH(vap)

where dH(vap) is latent heat of evaporization of water which is about 2,26 kJ/kg which you need to take in to account because water vapor turns to liquid water and that energy is released...
and now you enter the values in next equation and solve for temperature and there is your answer...

m1 cp1 (T2-T1) = m2 cp2 (T2-T1) + dH(vap)

where T2 is final temperature and T1 is starting temperature of each substance, so you solve that equation for T2 and T1 for copper is 0 °C and T1 for water is 100 °C

any questions?

anonymous · Answer

And how would you find the total entropy change of the system?