A 25 kg iron block initially at 280°C is quenched in an insulated tank that contains 100 kg of water at 18°C. Assuming the water that vaporizes during the process condenses back in the tank, determine the total entropy change during this process

During this process condenses back in the tank, calculate the total entropy change during this process. The specific heat of water at 25°C is cp = 4.18 kJ/kgK. The specific heat of iron at room temperature is cp = 0.45 kJ/kgK.

Mass of the iron block = m1 = 25kg

Initial temperature of the block = T1 = 280°C

Weight of the water = mw = 100kg

Temperature of water = Tw = 18°C

Specific heat of iron = cpI = 0.45kJ/kgK

Specific heat of water = cpw = 4.18kJ/kgK.

Energy balance equation:

ΔUI = −ΔUw

Here ΔUI is the internal energy of iron and ΔUw is the internal energy of water.

[mcp (T2 − T1)]I = − [mcp (T2 − T1)]w

cp is the specific heat at constant pressure.

Putting the values in above equation:

[25× 0.45(T2 − 280)]I = − [100 × 4.18(T2 − 18)]w

[11.25(T2 − 280)] = − [418(T2 − 18)]

11.25T2 − 3150 = − [418T2 − 7524]

11.25T2 − 3150 = −418T2 + 7524

11.25T2 + 418T2 = 7524 + 3150

429.25T2 = 10674

T2 = 24.86K

Calculate the total entropy change:

ΔStotal = ΔSI + ΔSw

ΔStotal = [mcp ln(T2/T1)]I + [mcp ln(T2/T1)]w

Now, we substitute the values in above equation:

ΔStotal = [25 × 0.45 ln(297.6/553)]I + [100 × 4.18 ln(297.6/291)]w

ΔStotal = −6.97 + 9.09

ΔStotal = 2.12kJ/K

The total entropy change is 2.12kJ/K

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