Answer:
There are 3 steps of this problem.
Explanation:
Step 1.
Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.
Step 2.
Enthalpy of saturated liquid Haq = 781.124 J/g
Enthalpy of saturated vapour Hvap = 2779.7 J/g
Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g
Step 3.
In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy
So, H1=H2
H2= (1-x)Haq+XHvap.........1
Putting the values in 1
2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}
= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)
1904.976 (J/g) = x1998.576 (J/g)
x = 1904.976 (J/g)/1998.576 (J/g)
x = 0.953
So, the quality of the wet steam is 0.953
Answer: Option D) covalent bonds between water molecules
In water, hydrogen bonds are best described as covalent bonds between water molecules
Explanation:
The hydrogen bonds between water molecules are covalent bonds because they are formed when oxygen attract the lone electron in hydrogen, thus resulting in the formation of a partially negative charge on the oxygen atom and a partially positive charge on two hydrogen atoms
Thus, the sharing of electrons between oxygen and hydrogen atoms is responsible for the covalent bonds between water molecules
Answer:
it is a replacement reaction
