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.

Answer: water at 18°C,  w cw = 4.18 kJ/kg∙°C; at 27oC, Cfe = 0.554 kJ/kg∙°C. (Note that we could also have the following units for the heat capacities: kJ/kg∙K.) Assuming that the heat capacities are constant we have the following

The final temperature of the iron and the water will be the same: Tfe, 2 = Tw, 2 = T2. Substituting T2 for Tfe, 2 and Tw, 2, and solving for T2 gives the following result for the final temperature.

T2 = ((Mw. Cw. T1w) + (Mfe. Cfe. T1fe)) / ((Mw. Cw) + (Mfe. Cfe)) ... equ 1

Where Me = mass of water = 100kg, Mfe = mass of iron = 25kg, T1fe = temp of iron before = 280°c,

Using substitution

T2 = ((100*4.18*18) + (25*0.554*280)) / ((100*4.18) + (25*.554))

T2 = 26.4°c

So determine total entropy change

DStot = DSw + DSfe ... equat3

DStot = final entropy, DSw = entropy of water at T2, DSfe = final entropy of iron at T2 where,

DS = M. C. lin (T2/T1) ... equ 5 temp is in Kelvin.

DSw = 100*4.18*lin (299.4/291) = 11.895

DSfe = 25*.554*lin (299.4/553) = - 8.498.

Substituting answers into equa3

DStot = 11.895 - 8.498 = 3.397kj/kg*Kelvin

Explanation: the explanation is in the answers above ...