Respuesta :
Answer:
The answer to the question is;
The total entropy change during this process is 4.03 kJ/K.
Explanation:
To solve the question, we first calculate the heat transferred from the iron to the water as follows
ΔU(iron) + ΔU(water) = 0
That is
m₁·Cp₁·(T₂ - 350) + m₂·Cp₂·(T₂ - 18) = 0
Where
Cp₁ = Specific heat capacity of iron = 0.444 J/g·°C
Cp₂ = Specific heat capacity of water = 4.186 J/g·°C
T₂ = final temperature and
m₁ = mass of iron = 25 kg
m₂ = mass of water = 100 kg
Substituting the values gives
25 kg × 0.444 J/g·°C × (T₂ - 350 °C) + 100 kg × 4.186 J/g·°C × (T₂ - 18 °C) = 0
11100 J/°C × (T₂ - 350 °C) + 418600 J/°C × (T₂ - 18 °C) = 0
11100 ×T₂ - 3885000 J +418600×T₂- 7534800 J = 0
429700 J/°C ×T₂ = 11419800 J
T₂ = 11419800 J÷ 429700 J/°C = 26.576 °C
Therefore the change in entropy is given by
For iron, change in entropy, ΔS₁ = m₁Cp₁㏑[tex](\frac{T_{21}}{T_{11}} )[/tex] = 25 kg×0.444 J/g·°C㏑[tex](\frac{299.726}{623.15} )[/tex] = -8124.296 J/K = -8.124 kJ/K
For water, change in entropy, ΔS₂ = m₂Cp₂㏑[tex](\frac{T_{22}}{T_{12}} )[/tex] = 100 kg×4.186 J/g·°C㏑[tex](\frac{299.726}{291.15} )[/tex] = 12152.01 J/K = 12.15201 kJ/K
Therefore the total change in entropy will be given by
ΔS[tex]_{tot}[/tex] = ΔS₁ + ΔS₂ = -8.124 kJ/K + 12.15201 kJ/K = 4.03 kJ/K.
The total entropy change during this process is 4.03 kJ/K this can be calculated using heat change formula.
Heat change:
ΔU(iron) + ΔU(water) = 0
m₁·Cp₁·(T₂ - 350) + m₂·Cp₂·(T₂ - 18) = 0
Where,
Cp₁ = Specific heat capacity of iron = 0.444 J/g·°C
Cp₂ = Specific heat capacity of water = 4.186 J/g·°C
T₂ = final temperature and
m₁ = mass of iron = 25 kg
m₂ = mass of water = 100 kg
On substituting the values gives
25 kg × 0.444 J/g·°C × (T₂ - 350 °C) + 100 kg × 4.186 J/g·°C × (T₂ - 18 °C) = 0
11100 J/°C × (T₂ - 350 °C) + 418600 J/°C × (T₂ - 18 °C) = 0
11100 ×T₂ - 3885000 J +418600×T₂- 7534800 J = 0
429700 J/°C ×T₂ = 11419800 J
T₂ = 11419800 J÷ 429700 J/°C = 26.576 °C
Calculation for change in entropy:
For iron, change in entropy, ΔS₁ = m₁Cp₁㏑ = 25 kg×0.444 J/g·°C㏑ = -8124.296 J/K = -8.124 kJ/K
For water, change in entropy, ΔS₂ = m₂Cp₂㏑ = 100 kg×4.186 J/g·°C㏑ = 12152.01 J/K = 12.15201 kJ/K
Therefore the total change in entropy will be given by
ΔS = ΔS₁ + ΔS₂ = -8.124 kJ/K + 12.15201 kJ/K = 4.03 kJ/K.
Find more information about Heat change here:
brainly.com/question/18912282
