2 lb_m of 20 wt H_2SO_4 aqueous solution at 100 degree F is mixed with 0.5 lb_m

Question

2 lb_m of 20 wt H_2SO_4 aqueous solution at 100 degree F is mixed with 0.5 lb_m of 80 wt % H_2SO_4 aqueous solution also at 100 degree F. (i) What is the composition and the temperature of the outlet mixture if you consider an adiabatic mixing? (ii) How much heat is to be removed or added to carry this mixing isothermally? (iii) How much would the final enthalpy of the mixture be if it behaved as an ideal solution when the two streams are mixed? For full credit you need to write down all H values you obtain from the H-x diagram, not just your final answers to the above questions.

Explanation / Answer

1 lb = 453.59 grams

so, 2 lb = 907.18 grams & 0.5 lb = 226.79 grams

1) 20 wt% H2SO4 so H2SO4 in soln is = 0.2 * 907.18 = 181.436 gram H2SO4

& water in soln .is = 0.8 * 907.18 = 725.74 gram water

2)80 wt% H2SO4 so H2SO4 in soln is = 0.8 * 226.79 = 181.432 gram H2SO4

& water in soln .is = 0.2 * 226.79 = 45.358 gram water

composition at outlet stream is,

1) + 2) = 181.436 + 181.432 = 362.868 gram H2SO4 & 725.74+45.358 = 771.098 gram water

TOTAL H2SO4 at Outlet in % = 362.868/ (362.868+771.098) = 362.868/1133.966 = 0.32 = 32 wt% H2SO4 & water at Outlet in % = 771.098/ (362.868+771.098) = 771.098/1133.966 = 0.68 = 68 wt% water & Temp = 100 F.

3) there is a liberation of heat always.

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