A lump of copper having a mass of 10 gm at a temperature of 500oC is just dropped into a well insulated bucket containing 100 gm of water at a temperature of 50oC. 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
[mCP(T1 - T2)]Copper = [mCP(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 = 54.2oC
Entropy change is given by
Δ
S = m CP ln (T2/T1)Where m is mass; CP is specific heat; T1 and T2 are initial and final temperatures respectively.
Entropy change of copper:
T1 = 273 + 500 = 773 K
T2 = 273 + 54.2 = 327.2 K
Δ
SCopper = 10 x 0.095 x ln (327.2/773) = -0.817 cal/oCEntropy change of water:
T1 = 273 + 50 = 323 K
T2 = 273 + 54.2 = 327.2 K
Δ
SWater = 100 x 1 x ln (327.2/323) = 1.292 cal/oCTotal entropy change
= ΔSCopper + ΔSWater = -0.817 + 1.292 = 0.475 cal/oCLast Modified on: 04-Feb-2022
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