Respuesta :
Answer:
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;
[tex]\int\limits^{T_f}_{273} {c_p} \, dT= -\int\limits^{T_f}_{373} {c_p} \, dT[/tex]
[tex]\int\limits^{T_f}_{273} {22.64 + 6.28 \times 10^{-3}\cdot T} \, dT= -\int\limits^{T_f}_{373} {22.64 + 6.28 \times 10^{-3}\cdot T} \, dT[/tex]
22.64·([tex]T_f[/tex] - 273) + (6.28 × 10⁻³)/2×([tex]T^2_f[/tex] - 273²) = - 22.64·([tex]T_f[/tex] - 373) - (6.28 × 10⁻³)/2×([tex]T^2_f[/tex] - 373²)
Which gives;
0.00314 × [tex]T^2_f[/tex] + 22.64 × [tex]T_f[/tex] - 6414.74 = -0.00314 × [tex]T^2_f[/tex] - 22.64 × [tex]T_f[/tex] + 8881.59
Transferring the right hand side of the equation to the left and collecting the like terms gives approximately;
0.00628× [tex]T^2_f[/tex] + 45.28 × [tex]T_f[/tex] - 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;
[tex]\Delta S_1 = \dfrac{\Delta Q}{T} =\int\limits^{323.318}_{273} { \left (\dfrac{22.64}{T} + 6.28 \times 10^{-3} \right)} \, dT\\\\\\\Delta S_1 = 22.64 \times ln \left ( \dfrac{323.318}{273} \right ) + 6.28 \times 10^{-3} \times (323.218 - 273) = 4.146 \ J/K[/tex]
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;
[tex]\Delta S_2 = \dfrac{\Delta Q}{T} =\int\limits^{323.318}_{373} { \left (\dfrac{22.64}{T} + 6.28 \times 10^{-3} \right)} \, dT\\\\\\\Delta S_2 = 22.64 \times ln \left ( \dfrac{323.318}{373} \right ) + 6.28 \times 10^{-3} \times (323.218 - 373) = -3.548 \ J/K[/tex]
The entropy introduced in the second mole of copper at 100°C, ΔS₂ = -3.548 J/K.
