One mole of copper at a uniform temperature of 0° C is placed in thermal contact with a second mole of copper which, initially, is at a uniform temperature of 100° C. Calculate the temperature of the 2-mole system, which is contained in an adiabatic enclosure, when thermal equilibrium is attained. Why is the common uniform temperature not exactly 50° C? How much thermal energy is transferred, and how much entropy is produced by the transfer? The constantpressure molar heat capacity of solid copper varies with temperature as cp = 22.64 + 6.28×10−3T J/mole K

1) T ≈ 50.318°C

2) It is due to the variation of the molar heat capacity of solid copper with temperature

3) The entropy introduced in the one mole of copper at 0°C is 4.146 J/K

The entropy introduced in the second mole of copper at 100°C is -3.548 J/K.

Explanation:

1) The given parameters are;

The temperature of 1 mole of copper = 0°C

The temperature of the second mole of copper 100°C

The specific heat capacity of copper = 22.64 + 6.28 × 10⁻³·T J/(mole·K)

The heat gained by the copper at 0° = The heat lost by the one mole of copper at 100°C

The total heat gained and lost is given as follows;

22.64·( - 273) +  (6.28 × 10⁻³)/2×( - 273²) = - 22.64·( - 373) -  (6.28 × 10⁻³)/2×(  - 373²)

Which gives;

0.00314 × + 22.64 × - 6414.74 = -0.00314 × - 22.64 × + 8881.59

Transferring the right hand side of the equation to the left and collecting the like terms gives approximately;

0.00628× + 45.28 × - 15296.33 = 0

Factoring with a graphing calculator, gives;

(T + 7533.509)·(T - 323.318)×0.00628 = 0

∴ T = 323.318 K ≈ 50.318°C

2) The common uniform temperature is not exactly 50°C because the hotter second mole of copper had a higher specific heat capacity, which resulted in the temperature at equilibrium being slightly higher than the average temperature

3) The amount of thermal energy transferred is given as follows;

22.64·(323.318 - 273) +  (6.28 × 10⁻³)/2×(323.318² - 273²) ≈ 1233.42 J

4) For the mole at 0°C, we have;

The entropy introduced in the one mole of copper at 0°C, ΔS₁ = 4.146 J/K

For the mole at 100°C, we have;

The entropy introduced in the second mole of copper at 100°C, ΔS₂ = -3.548 J/K.

1 is a answer of a question