In: Chemistry
Suppose that 0.36 g of water at 25 ∘C condenses on the surface of a 51-g block of aluminum that is initially at 25 ∘C. If the heat released during condensation goes only toward heating the metal, what is the final temperature (in ∘C) of the metal block? (The specific heat capacity of aluminum is 0.903 J/g∘C and the heat of vaporization of water at 25 ∘C is 44.0 kJ/mol.)
Solution :-
Lets find the amount of heat given by the condensation
mass of vapor = 0.36 g
heat of vaporization = 44.0 kJ/mol
q = heat of vap * mass of vap / molar mass of steam
0.36 g water * 44.0 kJ per mol / 18.016 g = 0.8792 kJ
0.8792 kJ * 1000 J / 1 kJ = 879.2 J
Now lets calculate the change in the temperature of the metal
q= m*c*deltaT
q/m*c = deltaT
879.2 J / 51 g * 0.903 J per g C = delta T
19.1 C = delta T
Final temperature of metal block = initial temperature + delta T
= 25.0 C + 19.1 C
= 44.1 C
so the final temperaure of the metal block is 44.1 oC