Answer to Question #86916 in General Chemistry for mike

Question

A piece of solid lead weighing 27.6 g at a temperature of 314 °C is placed in 276 g of liquid lead at a temperature of 370. °C. After a while, the solid melts and a completely liquid sample remains. Calculate the temperature after thermal equilibrium is reached, assuming no heat loss to the surroundings.

The enthalpy of fusion of solid lead is ΔHfus = 4.77 kJ/mol at its melting point of 328 °C, and the molar heat capacities for solid and liquid lead are Csolid = 26.9 J/mol K and Cliquid = 28.6 J/mol K.

Tfinal = ______ °C

Answer 1

n_sol(Pb) = m_sol(Pb) / M(Pb) = 27.6 g / 207 g/mol = 0.1333 mol.

n_liq(Pb) = m_liq(Pb) / M(Pb) = 276 g / 207 g/mol = 1.3333 mol.

T_sol1(Pb) = t_sol1(Pb) + 273.15 = (314 + 273.15) K = 587.15 K.

T_liq1(Pb) = t_liq1(Pb) + 273.15 = (370 + 273.15) K = 643.15 K.

T_melt(Pb) = t_melt(Pb) + 273.15 = (328 + 273.15) K = 601.15 K.

The temperature after thermal equilibrium is T_final(Pb) = X.

Q₁ = C_liq(Pb) × n_liq(Pb) × (T_liq1(Pb) –T_final(Pb)) = 28.6 J/mol K × 1.3333 mol × (643.15K – T_final(Pb)).

Q₂ = C_sol(Pb) × n_sol(Pb) × (T_melt(Pb) – T_sol1(Pb)) + ΔH_fus(Pb) × n_sol(Pb) + C_liq(Pb) × n_sol(Pb) × (T_final(Pb) – T_melt(Pb)) = 26.9 J/mol K × 0.1333 mol × (601.15K – 587.15K) + 4770 J/mol × 0.1333 mol + 28.6 J/mol K × 0.1333 mol × (T_final(Pb) – 601.15 K).

Q₁ = Q₂

28.6 × 1.3333 × (643.15 – X) = 26.9 × 0.1333 mol × (601.15 – 587.15) + 4770 × 0.1333 + 28.6 × 0.1333 × (X – 601.15)

24524.84 – 38.13X = 50.20 + 635.84 + 3.81X – 2291.81

41.94X = 26130.61

X = 623.05. T_final(Pb) = 623.05K.

t_final(Pb) = T_final(Pb) – 273.15 = (623.05 – 273.15)⁰C = 349.90 ⁰C ≈ 350⁰C.