Question

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.)​

Answer 1

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