Thermal Equilibrium: A block of cast iron of mass 10 kg is heat treated at 800 K

Question

Thermal Equilibrium: A block of cast iron of mass 10 kg is heat treated at 800 K and dropped into a 95 kg water bath, initially at 300 K, to cool. Both the cast iron and water can be assumed to have constant specific heats capacities. Assuming that the two materials together are contained in an insulated container that allows no heat transfer to the surroundings, determine the final thermal equilibrium temperature of the block and water, T2. Also calculate the change in entropy of the block, SFE, the change in entropy of the water, SW, and the total change in entropy of the system, Stot.

Explanation / Answer

Heat lost by block = heat gained by water

let Tf be the final temperature

10 x 1000 x 0.450 x (800 - Tf) = 95 x 1000 x 4.18 x (Tf - 300)

3600 - 4.5Tf = 397.1Tf - 119130

401.6Tf = 122730

Tf = 305.60 K

Change in entropy of block dSfe = 10 x 1000 x 0.45 x 494.4/494.4 = 4500 J/K

change in entropu of water dSw = 95 x 1000 x 4.18 x 5.60/5.60 = 397100 J/K

Total entropy dStot = 401700 J/K

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.