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:
For - 5.556 lb/s
For - 7.4047 lb/s
Solution:
As per the question:
System Load = 96000 Btuh
Temperature, T =
Temperature rise, T' =
Now,
The system load is taken to be at constant pressure, then:
Specific heat of air,
Now, for a rise of in temeprature:
Now, for :
Answer:
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Answer:
fluid nozzle that is too large