Answer: ALL CAREFULLY ANSWERED CORRECTLY.
Explanation:
1) A loaf of Bread PHYSICAL SYSTEM
✓ How can the environment affect the edibility of the bread
✓ What are the constituents that makes up the bread
✓ What process is involved in these constituents mixing to form the loaf.
2) The law of thermodynamics makes us to understand that when heat/energy passes through a system, the systems internal energy changes with respect to the conservation of energy law. That is energy lost = energy gained. Typically, ice would melt in a cup of hot tea because of the thermal energy in the molecules of the hot tea. When you heat a material, you are adding thermal kinetic energy to its molecules and usually raising its temperature. The temperature of the ice raises due to the kinetic energy added to it and it melts to water.
3) The theory of systems view the world as a complex system of interconnected parts. If we consider the society; (financial systems, political systems, etc) we will agree that they individually have their own components and it's the summation of this components that makes the system, this implies that system thinking could be applicable in this kinda of systems as long as they are made up of components.
4) Technology has boosted every sector of our lives and it has the capacity to do more. Restricting it's importance to entertainment alone would be an underusing of its potentials. Engineering students infact should not need any drive to be encouraged about maximizing all it can do in shaping our world.
5) ~ Nature shows its splendid soul
~Never ceases to leave us in amazement
~And we are in love
Answer:
A stack is an ordered list of elements where all insertions and deletions are made at the same end, whereas a queue is exactly the opposite of a stack.
Explanation:
Answer:
final temperature is 424.8 K
so correct option is e 424.8 K
Explanation:
given data
pressure p1 = 1 bar
pressure p2 = 5 bar
index k = 1.3
temperature t1 = 20°C = 293 k
to find out
final temperature t2
solution
we have given compression is reversible and has an index k
so we can say temperature is
![\frac{t2}{t1}= [\frac{p2}{p1}]^{\frac{k-1}{k} }](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7Bt1%7D%3D%20%5B%5Cfrac%7Bp2%7D%7Bp1%7D%5D%5E%7B%5Cfrac%7Bk-1%7D%7Bk%7D%20%7D)
...........1
put here all these value and we get t2
![\frac{t2}{293}= [\frac{5}{1}]^{\frac{1.3-1}{1.3} }](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7B293%7D%3D%20%5B%5Cfrac%7B5%7D%7B1%7D%5D%5E%7B%5Cfrac%7B1.3-1%7D%7B1.3%7D%20%7D)
t2 = 424.8
final temperature is 424.8 K
so correct option is e
Answer:
576.21kJ
Explanation:
#We know that:
The balance mass 
so, 
#Also, given the properties of water as;
#We assume constant properties for the steam at average temperatures:
#Replace known values in the equation above;
#Using the mass and energy balance relations;
#We have 
: we replace the known values in the equation as;
#Hence,the amount of heat transferred when the steam temperature reaches 500°C is 576.21kJ
Answer:
A.) P = 2bar, W = - 12kJ
B.) P = 0.8 bar, W = - 7.3 kJ
C.) P = 0.608 bar, W = - 6.4kJ
Explanation: Given that the relation between pressure and volume is
PV^n = constant.
That is, P1V1^n = P2V2^n
P1 = P2 × ( V2/V1 )^n
If the initial volume V1 = 0.1 m3,
the final volume V2 = 0.04 m3, and
the final pressure P2 = 2 bar.
A.) When n = 0
Substitute all the parameters into the formula
(V2/V1)^0 = 1
Therefore, P2 = P1 = 2 bar
Work = ∫ PdV = constant × dV
Work = 2 × 10^5 × [ 0.04 - 0.1 ]
Work = 200000 × - 0.06
Work = - 12000J
Work = - 12 kJ
B.) When n = 1
P1 = 2 × (0.04/0.1)^1
P1 = 2 × 0.4 = 0.8 bar
Work = ∫ PdV = constant × ∫dV/V
Work = P1V1 × ln ( V2/V1 )
Work = 0.8 ×10^5 × 0.1 × ln 0.4
Work = - 7330.3J
Work = -7.33 kJ
C.) When n = 1.3
P1 = 2 × (0.04/0.1)^1.3
P1 = 0.6077 bar
Work = ∫ PdV
Work = (P2V2 - P1V1)/ ( 1 - 1.3 )
Work = (2×10^5×0.04) - (0.608 10^5×0.1)/ ( 1 - 1.3 )
Work = (8000 - 6080)/ -0.3
Work = -1920/0.3
Work = -6400 J
Work = -6.4 kJ
