Question

An iron cube at a temperature of 400°C is dropped into an insulated bath containing 10 kg water at 25°C. The water finally reaches a temperature of 50°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 ?

Answer 1

Given : Temperature of iron cube = 400°C = 673 K 

Temperature of water = 25°C = 298 K 

Mass of water = 10 kg 

Temperature of water and cube after equilibrium = 50°C = 323 K 

Specific heat of water, c_pw = 4186 J/kg K 

Entropy changes for the iron cube and the water : 

Is the process reversible ? 

Now, Heat lost by iron cube = Heat gained by water 

m_i c_pi (673 – 323) = m_w c_pw (323 – 298) 

= 10 × 4186 (323 – 298)

Entropy of iron at 673 K

Entropy of water at 298 K

Entropy of iron at 323 K = 2990 × ln \(\left(\cfrac{323}{273}\right)\) = 502.8 J/K

Entropy water at 323 K = 10 × 4186 ln \(\left(\cfrac{323}{273}\right)\) = 7040.04 J/K

Changes in entropy of iron = 502.8 – 2697.8 = – 2195 J/K 

Change in entropy of water = 7040.04 – 3667.8 = 3372.24 J/K 

Net change in entropy = 3372.24 – 2195 = 1177.24 J/K 

Since ∆S > 0 hence the process is irrevesible.