25 mole of solid copper initially at 1200 K are quenched in 200 mows of liquid w

Question

25 mole of solid copper initially at 1200 K are quenched in 200 mows of liquid water (H_2O) initially at 300 K. The total process occurs within a closed adiabatic system at constant pressure. No liquid water is lost to evaporation. (a) Calculate the final system temperature. (b) Calculate Delta H (extensive) (c) Calculate Delta S consider enthaply: H = H(T, V) (a) Derive: partial differential H/partial differential T in terms of properties selected from the set P, T, V, alpha k, G_p and C_V you can use any equations from HO3 and 4, but you MUST show your work. (b) Derive: partial differential H/partial differential V in terms of properties selected from the set P, T, V, alha, k, C_p and C_V

Explanation / Answer

1a) Let the constant temperature be T K.

Moles of Cu (s), n1 = 25.00

Moles of H2O (l), n2 = 200

Initial temperature of Cu (s), T1 = 1200 K

Initial temperature of H2O (L), T2 = 300 K

cp (Cu) = 34 J/(mole K)

cp (H2O) = 75.4 J/(mole K)

Employ the principle of thermochemistry: Heat lost by hot body = heat gained by cold body.

Heat lost by Cu = (25 moles)*(34 J/mol.K)*(1200 – T) K = 850*(1200 – T)

Heat gained by H2O = (200 moles)*(75.4 J/mol.K)*(T – 300) K = 15080*(T – 300)

Therefore,

850*(1200 – T) = 15080*(T – 300)

===> 1020000 – 850T = 15080T – 4524000

===> 1020000 + 4524000 = 15080T + 850T

===> 5544000 = 15930T

===. T = 5544000/15930 = 348.0226 348.02

The final system temperature is 348.02 K (ans).

b) H (extensive, Cu system) = (25 moles)*(34 J/mol.K)*(1200 – 348.02) K = 724183 J = 724.183 kJ (ans).

H (extensive, H2O system) = (200 moles)*(75.4 J/mol.K)*(348.02 – 300) K = 724141.6 J = 724.1416 kJ 724.142 kJ (ans).

H (extensive, system) = 0. The reason is that the heat lost by the hot Cu block is gained by the water and hence the total heat involved in the system is zero, i.e, no heat is lost or gained by the system. This is true if we realize that we are working with an adiabatic system and hence, the total heat involved must be zero.

c) S (extensive, Cu system) = 724183 J/348.02 K = 2080.866 J/K 2080.9 J/K (ans).

S (extensive, H2O system) = 724141.6 J/348.02 K = 2080.747 J/K 2080.7 J/K (ans).

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