A 60 kg iron block initially at 70 C js dropped into a lake at 20 C.)Thermal equ

Question

A 60 kg iron block initially at 70 C js dropped into a lake at 20 C.)Thermal equilibrium is established after a while as a result of heat transfer between the block and the lake water. The specific heat of ipon at room temperature is op 0.45 kJ/kg c Assuming the surroundings to be at 17 C, a. Determine the internal energy change and the entropy change of the iron. b. Determine the total entropy change for this process. ?cedD " Determine the exergy at the initialtate. ?./ etermine the exergy change for the reversible process tXL . o)-To e 20 C Iron Fe

Explanation / Answer

a.

Energy change of iron = m * Cp * (T1 - T2)

= 60 * 0.45 * (70 - 20)

= 1350 kJ

Entropy change of iron = m Cp ln (T2 / T1)

= 60 * 0.45 * ln (20+273 / 70+273)

= -4.254 kJ/K

b)

Q - W = dU

since W = 0, we get Q = dU

Entropy change of lake = Q / T

= 1350 / (273 + 20)

= 4.6075 kJ/K

Total entropy change for process = Entropy change of iron + lake

= - 4.254 + 4.6075

= 0.3535 kJ/K

c)

Exergy change = Exergy destroyed = Tamb * Sgen

= (273+17) * 0.3535

= 102.52 kJ

d)

Exergy E = (U - U0) + P0 (v - v0) + - T0 (s - s0)

If iron were to come to ambient temperature, energy transfer u - u0 = m * Cp * (T1 - T0)

= 60 * 0.45 * (70 - 17)

= 1431 kJ

Exergy = 1431 + 102.52 = 1533.5 kJ

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