Entropy Change in Adiabatic Heat Exchange

A lump of copper having a mass of 10 gm at a temperature of 500^oC is just dropped into a well insulated bucket containing 100 gm of water at a temperature of 50^oC. If the heat capacities of copper and water are 0.095 and 1.0 respectively, expressed as cal/gm.^oC, calculate the total change in entropy resulting from the process.

Calculations:

Final temperature of copper and water will be the same, and is obtained by enthalpy balance.

Heat lost by copper = heat gained by water

[mC_P(T₁ - T₂)]_Copper = [mC_P(T₂ - T₁)]_Water

10 x 0.095 x (500 - T₂) = 100 x 1 x (T₂ - 50)

0.95 x (500 - T₂) = 100 x (T₂ - 50)

(500 - T₂) = 105.26 x (T₂ - 50)

500 + 5263.2 = 106.26 T₂

T₂ = 54.2^oC

Entropy change is given by

ΔS = m C_P ln (T₂/T₁)

Where m is mass; C_P is specific heat; T₁ and T₂ are initial and final temperatures respectively.

T₁ = 273 + 500 = 773 K

T₂ = 273 + 54.2 = 327.2 K

ΔS_Copper = 10 x 0.095 x ln (327.2/773) = -0.817 cal/^oC

T₁ = 273 + 50 = 323 K

T₂ = 273 + 54.2 = 327.2 K

ΔS_Water = 100 x 1 x ln (327.2/323) = 1.292 cal/^oC

Total entropy change = ΔS_Copper + ΔS_Water = -0.817 + 1.292 = 0.475 cal/^oC