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Entropy Change in Adiabatic Heat Exchange

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A lump of copper having a mass of 10 gm at a
temperature of 500^{o}C is just dropped into a well insulated bucket containing 100 gm of water at a temperature of 50^{o}C. If the heat capacities of copper and water are 0.095 and 1.0 respectively, expressed as cal/gm.^{o}C, 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}(T_{1} - T_{2})]_{Copper} = [mC_{P}(T_{2} - T_{1})]_{Water}

10 x 0.095 x (500 - T_{2}) = 100 x 1 x (T_{2} - 50)

0.95 x (500 - T_{2}) = 100 x (T_{2} - 50)

(500 - T_{2}) = 105.26 x (T_{2} - 50)

500 + 5263.2 = 106.26 T_{2}

T_{2} = 54.2^{o}C

Entropy change is given by

ΔS = m C_{P} ln (T_{2}/T_{1})

Where m is mass; C_{P} is specific heat; T_{1} and T_{2} are initial and final temperatures respectively.

Entropy change of copper:

T_{1} = 273 + 500 = 773 K

T_{2} = 273 + 54.2 = 327.2 K

ΔS_{Copper} = 10 x 0.095 x ln (327.2/773) = -0.817 cal/^{o}C

Entropy change of water:

T_{1} = 273 + 50 = 323 K

T_{2} = 273 + 54.2 = 327.2 K

ΔS_{Water} = 100 x 1 x ln (327.2/323) = 1.292 cal/^{o}C

Total entropy change = ΔS_{Copper }+ ΔS_{Water }= -0.817 + 1.292 = 0.475 cal/^{o}C

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Last Modified on: 01-May-2024

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