A 200-gram block of aluminum (Al, 27 g/mole) with an initial temperature T_Al =

Question

A 200-gram block of aluminum (Al, 27 g/mole) with an initial temperature T_Al = 76.85 degree C is brought into contact with a 500-gram block of copper (Cu, 63.5 g/mole) with an initial temperature T_Cu = 26.85 degree C. Together the two blocks reach a final equilibrium temperature T_F. Assume that the blocks only exchange heat with one another, and that the blocks do not change in volume. Let T_F be the final temperature calculated using the Einstein model of solids where each metal block can be treated as a three dimensional lattice of balls on springs. The drawing shows two dimensions of the 3-dimensional lattice. Atoms (red balls) oscillate around their nominal positions (blue dots). Springs represent the potential V = (k/2) (x^2 + y^2 + z^2). In this approximation, we find the heat capacity at constant volume using the equipartition theorem. What is the final temperature of the two blocks in this approximation?

Explanation / Answer

Mass of copper block = 500 g

Molar mass of copper = 63.5 g

Temperature of copper = 26.85oC = (26.85 + 273) K = 299.85 K

Mass of Aluminium block = 200 g

Molar mass of Aluminium = 27 g

Temperature of Aluminium = 76.85oC = (76.85 + 273) K = 348.85 K

Heat capacity = Cv = 3kB.

Final temperature of the blocks can be calculated by using the equation, Q = TiTfCvdT.

Let us first calculate the heat transferred from copper block to Aluminium block.

E Cu int = 3kBT = 3kB × 299.85 K = 899.95kB J/atom

E Al int = 3kBT = 3kB × 348.85 K = 1049kB J/atom.

Q goes from Al to Cu, because average (or per atom) internal energy in Al is higher, even though total internal energy in Cu is higher because it has more atoms. If Avogadro’s number is represented by NA, then,

E Cu int (total) = 500 g /63.5 g/mol × 899.9kB J/atom × NA atoms = 7086.22kBNA Joules

E Al int(total) = 200 g /27 g/mol × 879kB J/atom × NA atoms = 7770.37kBNA Joules.

A Aluminium atom loses energy xkB Joules, which is gained by an Cu atom.

Since energy is conserved,

1049kB xkB = 899.95kB + xkB x = 74.52.

Therefore 74.52Joules of energy are transferred.

Q = -74.52 kBJoules

But Q = TiTfCvdT.

Substituting the value of Q and Cv, we obtained,

-74.52 kB = TiTf3kBdT

= 3kB TiTfdT

= 3kB(Tf Ti)

= 3kB(Tf 348.85)

final temparaturw Tf = 324 K = 51 C.

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