Answer:
b. The pirating streams are eroding headwardly to intersect more of the other streams’ drainage basins, causing water to be diverted down their steeper gradients.
Explanation:
From the Kaaterskill NY 15 minute map (1906), this shows two classic examples of stream capture.
The Kaaterskill Creek flow down the east relatively steep slopes into the Hudson River Valley. While, the Gooseberry Creek is a low gradient stream flowing down the west direction which in turn drains the higher parts of the Catskills in this area.
However, there is Headward erosion of Kaaterskill Creek which resulted to the capture of part of the headwaters of Gooseberry Creek.
The evidence for this is the presence of "barbed" (enters at obtuse rather than acute angle) tributary which enters Kaaterskill Creek from South Lake which was once a part of the Gooseberry Creek drainage system.
It should be noted again, that there is drainage divide between the Gooseberry and Kaaterskill drainage systems (just to the left of the word Twilight) which is located in the center of the valley.
As it progresses, this divide will then move westward as Kaaterskill captures more and more of the Gooseberry system.
Answer:
The change in entropy is found to be 0.85244 KJ/k
Explanation:
In order to solve this question, we first need to find the ration of temperature for both state 1 and state 2. For that, we can use Charles' law. Because the volume of the tank is constant.
P1/T1 = P2/T2
T2/T1 = P2/P1
T2/T1 = 180 KPa/120KPa
T2/T1 = 1.5
Now, the change in entropy is given as:
ΔS = m(s2 - s1)
where,
s2 = Cv ln(T2/T1)
s1 = R ln(V2/V1)
ΔS = change in entropy
m = mass of CO2 = 3.2 kg
Therefore,
ΔS = m[Cv ln(T2/T1) - R ln(V2/V1)]
Since, V1 = V2, therefore,
ΔS = mCv ln(T2/T1)
Cv at 300 k for carbondioxide is 0.657 KJ/Kg.K
Therefore,
ΔS = (3.2 kg)(0.657 KJ/kg.k) ln(1.5)
<u>ΔS = 0.85244 KJ/k</u>
Answer:
Enthalpy of reaction (kJoules/mole)
Heat of formation of products (kJoules/mole)
Heat of reaction of reactants (kJoules/mole)
Explanation:
The general expression for calculating the overall enthalpy of reaction is given as following:
ΔH = ∑ΔH[producst] - ∑Δ[reactants]
Thus, the heat of reaction is given as the difference between the formation of the products and the formation of the reactants. The units are expressed as kJ/mol of reactants or products.
Thus, the three values are fundamental in the determination of the overall energy of the reaction from Hess' Law.
