Answer:
(a) ΔU = 125 kJ
(b) ΔU = -110 kJ
Explanation:
<em>(a) Suppose that 150 kJ of work are used to compress a spring, and that 25 kJ of heat are given off by the spring during this compression. What is the change in internal energy of the spring during the process?</em>
<em />
The work is done to the system so w = 150 kJ.
The heat is released by the system so q = -25 kJ.
The change in internal energy (ΔU) is:
ΔU = q + w
ΔU = -25 kJ + 150 kJ = 125 kJ
<em>(b) Suppose that 100 kJ of work is done by a motor, but it also gives off 10 kJ of heat while carrying out this work. What is the change in internal energy of the motor during the process?</em>
<em />
The work is done by the system so w = -100 kJ.
The heat is released by the system so q = -10 kJ.
The change in internal energy (ΔU) is:
ΔU = q + w
ΔU = -10 kJ - 100 kJ = -110 kJ
Answer:
Enthalpy at outlet=284.44 KJ
Explanation:


We need to Find enthalpy of outlet.
Lets take the outlet mass m and outlet enthalpy h.
So from mass conservation

m=1+1.5+2 Kg/s
m=4.5 Kg/s
Now from energy conservation

By putting the values

So h=284.44 KJ
Answer:
1964
Explanation:
It was discovered in 1964 when a pair of Geiger counters were carried on board a sub-orbital rocket launched from New Mexico.
Answer:
C. Form follows function.