What is the temperature of a water body containing a 90% saturation and 11.5 ppm of dissolved oxygen?

Using the chart that has been provided, we may determine water temperature. We do this by drawing a straight line form the bottom scale which has the ppm of oxygen dissolved to the middle scale which has the percentage saturation.

The line starts from 11.5 ppm on the bottom scale and goes to 90% on the middle scale. Next, we continue this line, without changing its slope, to the third scale showing temperature. We see that it crosses the temperature scale at 4°C.

The temperature of the water is 4 °C.

Answerspecies;

Given data,temperature, t = 230°c = 503 kconsider the chemical reaction,ch₃oh(l) + 1/2 o₂(g)  → ch₂o(g) + h₂o(l)calculate the enthalpy change for the reaction: δh⁰rxn =  σnδh⁰f(products) -  σmδh⁰f(reactants)substitute the values from the table: δh⁰rxn = { -115.9 + (-285.8)} - {0+(-238.7)}              = -163.0 kjcalculate the entropy change for the reaction.δs⁰rxn = (219 + 70) - (127.2 + 1/2 x 205.2) = 59.2 j/kconsider the expression for free energy: δg⁰rxn =  δh⁰rxn - t  δs⁰rxnsubstitute the values in the above equationδg⁰rxn = -163 x 10³ - (503 x 59.2)              = -192777.6 j              = -192.777 kj