Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Answer:
0.05 J/K
Explanation:
Given data in question
heat (Q) = 10 J
temperature (T) = 200 K
to find out
the change in entropy of the system
Solution
we will solve this by the entropy change equation
i.e ΔS = ΔQ/T ...................1
put the value of heat Q and Temperature T in equation 1
ΔS is the enthalpy change and T is the temperature
so ΔS = 10/200
ΔS = 0.05 J/K
Answer:
increases by a factor of 6.
Explanation:
Let us assume that the initial cross sectional area of the pipe is A m² while the initial velocity of the water is V m/s², hence the flow rate of the water is:
Initial flow rate = area * velocity = A * V = AV m³/s
The water speed doubles (2V m/s) and the cross-sectional area of the pipe triples (3A m²), hence the volume flow rate becomes:
Final flow rate = 2V * 3A = 6AV m³/s = 6 * initial flow rate
Hence, the volume flow rate of the water passing through it increases by a factor of 6.
Moisture content is measured in terms of pounds of water per pound of air (lb water/lb air) or grains of water per pound of air (gr. of water/lb air).
Hope this helps❤
