The alkaline earth metals (the second group) because their ion charge is +2
You can determine it by paying attention to the unique characteristics that could only be found at heart's tissue, such as :
- looks striated or stripped
- The bundles are breached like tree but connected at both ends
hope this helps
Answer:
The final temperature is 39.58 degree Celsius
Explanation:
As we know
Q = m * c * change in temperature
Specific heat of water (c) = 4.2 joules per gram per Celsius degree
Substituting the given values we get -
5750 = 335 * 4.2 * (X - 35.5)
X = 39.58 degree Celsius
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
