Question

A cooper block having a mass of 10 kg and at a temperature of 800 K is placed in a well-insulated vessel containing 100 kg of water initially at 290 K. Calculate:

a) Calculate the entropy change for the block, the water, and the total process.

b) What is the maximum amount of work that could have been obtained from the copper block and water in a Carnot engine? The heat capacities are 4.185 kJ/kg/K for water and 0.398 kJ/kg/K for copper.

Answer 1

Heat gained by water = heat released by copper

Mass x Cp x (T -T1) = mass of Cu x Cp x (T2-T)

100 kg x 4.185 kJ/kg-K x (T - 290)K = 10 kg x 0.398 kJ/kg-K x (800 - T)K

105.1507 (T - 290) = 800 - T

105.1507 T - 30493.72 = 800 - T

T = 294.8 K

Part a

entropy change for the block

S1 = mCp ln (T/T2)

= 10 kg x 0.398 kJ/kg-K x ln (294.8/800)

= - 3.973 kJ/K

entropy change for the water

S2 = mCp ln (T/T1)

= 100 kg x 4.185 kJ/kg-K x ln (294.8/290)

= 6.8701 kJ/K

Entropy change of process = S1 + S2

= - 3.973 + 6.8701

= 2.897 kJ/K

Part b

maximum amount of work can be obtained when efficiency is maximum

Efficiency of Carnot engine = 100% = 1

Maximum work obtained = efficiency x Qh

= 1 x 100 kg x 4.185 kJ/kg-K x (294.8 - 290)K

= 2008.8 kJ