Question

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 1

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..