A 25-kg iron block initially at 350 oC is quenched in an insulated tank that contains 100 kg of water at 18 oC. Assuming the water that vaporizes during the process condenses back in the tank, determine the total entropy change during this process.

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㏑ = 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.

where is the following

explanation:

no, the red giant is the hottest with 2.00.000.000 degrees celsius. while the white dwarf only 25,000 degrees celsius.