An iron cube at a temperature of 400^o C is dropped into an insulated bath containing 10kg water at 25^o C. the water finally reaches a temperature of 50^o C at steady state. Given that the specific heat of water is equal to 4186 J / kg K. find the entropy changes for the iron cube and the water. Is the process reversible? If so why?

Given: temperature of the iron cube = 400^o C = 400 + 273 = 673 K

Temperature of water = 10 kg
Temperature of water and cube after equilibrium = 50^o C = 50 +273 = 323 K
Specific heat of water, c_pw = 4186 J/kg K
Entropy changes for the iron cube and the water:
Heat lost by iron cube = heat gained by water
Therefore, m_i c_pi (673 – 323) = m_w c_pw (323 – 298)
= 10 X 4186 (323 – 298)
m_i c_pi = 10 X 4186 (323 – 298) / (673 – 323) = 2990
Where m_i = mass of iron (kg)
c_pi = specific heat of iron (J/kg K)
Entropy of iron at 673 K = m_i c_pi ln (673 / 273) = 2990 ln (673 / 273)
= 2697.8 J/K

Entropy of water at 298 K = m_w c_pw ln (298 / 273)
= 10 X 4186 ln (298 / 273) = 3667.8 J/K
Entropy of iron at 323 K = 2990 X ln (323 / 273) = 502.8 J/K
Entropy of water at 323 K = 10 X 4186 ln (323 / 273) = 7040.04 J/K
Change in entropy of iron = 502.8 – 2697.8 = – 2195 J/K
Change in entropy of water = 7040.04 – 3667.8 = 3382.24 J/K
Net change in entropy = 3372.34 – 2195 = 1177.24 J/K
Since ∆S >0, hence the process is irreversible