Question

. Найти изменение энтропии при плавлении 2 кг. свинца и дальнейшем его охлаждении от 327 до 0˚С.

Answer 1

1. Плавление
dQ=λ*m (2)
где m=2 кг, λ - удельная теплота плавления, для свинца λ=22,6 кДж/К
Процесс плавления происходит при постоянной температуре плавления Тпл=327 К, и подставляя (2) в (1), затем интегрируя, получаем:
ΔS=λ*m/Тпл=22,6*103*2/327=138,2 Дж/К
2. Охлаждение
dQ=c*m*dT (3)
где с - удельная теплоемкость, для свинца с=126 Дж/(кг*К)
Подставляя (3) в (1) и интегрируя в пределах от T1=600 K до T2=273 К получаем:
ΔS=с*m*ln(273/600)=126*2*ln(273/600)=-198,43 Дж/К

 

Answer 2

Calculation of Change in Entropy during Melting and Cooling of Lead

To calculate the change in entropy during the melting and subsequent cooling of 2 kg of lead from 327°C to 0°C, we need to consider the entropy changes at each stage.

1. Melting of Lead: - The entropy change during the melting process can be calculated using the formula: ΔS = Q/T, where ΔS is the change in entropy, Q is the heat transferred, and T is the temperature. - The heat transferred during the melting process can be calculated using the formula: Q = m * ΔH, where m is the mass of the substance and ΔH is the enthalpy of fusion. - The enthalpy of fusion for lead is approximately 24.5 kJ/mol. - The molar mass of lead is approximately 207.2 g/mol. - The number of moles of lead can be calculated using the formula: moles = mass / molar mass. - The temperature during the melting process is constant at the melting point of lead, which is 327°C.

2. Cooling of Lead: - The entropy change during the cooling process can be calculated using the formula: ΔS = Q/T, where ΔS is the change in entropy, Q is the heat transferred, and T is the temperature. - The heat transferred during the cooling process can be calculated using the formula: Q = m * C * ΔT, where m is the mass of the substance, C is the specific heat capacity, and ΔT is the change in temperature. - The specific heat capacity of lead is approximately 0.13 J/g°C. - The change in temperature during the cooling process is from 327°C to 0°C.

3. Calculation: - Let's calculate the change in entropy during the melting and cooling of 2 kg of lead.

- Melting of Lead: - The mass of lead is 2 kg, which is equal to 2000 g. - The number of moles of lead can be calculated as follows: - moles = mass / molar mass = 2000 g / 207.2 g/mol ≈ 9.64 mol. - The heat transferred during the melting process can be calculated as follows: - Q = m * ΔH = 9.64 mol * 24.5 kJ/mol = 235.78 kJ. - The change in entropy during the melting process can be calculated as follows: - ΔS = Q / T = 235.78 kJ / (327 + 273) K = 0.439 kJ/K.

- Cooling of Lead: - The mass of lead is 2 kg, which is equal to 2000 g. - The change in temperature during the cooling process is from 327°C to 0°C, which is equal to 327 K. - The heat transferred during the cooling process can be calculated as follows: - Q = m * C * ΔT = 2000 g * 0.13 J/g°C * 327 K = 85,020 J. - The change in entropy during the cooling process can be calculated as follows: - ΔS = Q / T = 85,020 J / 327 K = 260.06 J/K.

4. Total Change in Entropy: - The total change in entropy during the melting and cooling of 2 kg of lead is the sum of the entropy changes during the two processes. - ΔS_total = ΔS_melting + ΔS_cooling = 0.439 kJ/K + 260.06 J/K = 0.439 kJ/K + 0.26 kJ/K = 0.699 kJ/K.

Therefore, the change in entropy during the melting and subsequent cooling of 2 kg of lead from 327°C to 0°C is approximately 0.699 kJ/K.

Please note that the calculations provided are based on the given information and assumptions.