100 cm3 of solid copper initially at 1000o C are quenched in 3000 cm3 of liquid

Question

100 cm3 of solid copper initially at 1000o C are quenched in 3000 cm3 of liquid water (H2O) initially at T=25o C. 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(s)

(b) Calculate the change in H (extensive) for the copper subsystem and for the water subsystem.

Explanation / Answer

specific heat capacity of copper = 0.385 kJ/kg K specific heat capacity of water = 4.186kJ/kg K let the final temperature be T then mass of copper = 100*8.96 = 896g = 0.896kg mass of water = 3000*1 = 3000g = 3kg now a)heat exchanged = mass* specific heat*change in temperature =>0.896*0.385(T-1000) + 3*4.186*(T-25) = 0 =>T = 51 deg C b)change in H in water = 3*4.186(51-25) = 326.5KJ change in H in copper = -326.5

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