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HACTEHA [7]
4 years ago
15

Four kilograms of carbon monoxide (co) is contained in a rigid tank with a volume of 1 m3. the tank is fitted with a paddle whee

l that transfers energy to the co at a constant rate of 14 w for 1 h. during the process, the specific internal energy of the carbon monoxide increases by 10 kj/kg. if no overall changes in kinetic and potential energy occur, determine (a) the specific volume at the final state, in m3/kg. (b) the energy transfer by work, in kj. (c) the energy transfer by heat transfer, in kj, and the direction of the heat transfer.
Physics
1 answer:
Alenkasestr [34]4 years ago
6 0
Part (a): Specific volume
Specific volume, v = V/m, V = Volume of the tank, m = mass of CO in the tank
Therefore,
v = 1/4 = 0.25 m^3/kg

Part (b): Energy transferred in kJ
Work done, W = PΔt, where P = power = 14 W = 14 J/s, t = time = 1 hour = 60*60 = 3600 seconds
Therefore,
W = 14*3600 = 50400 J = 50.4 kJ

Part (c): Energy transferred by heat
ΔU = Q + W
Then,
Heat transferred by heat, Q = ΔU - W
But, ΔU = mΔu = 4 kg*10 kJ/kg = 40 kJ
Therefore,
Q = 40 - 50.4 = -10.4 kJ (negative sign indicates that heat is removed from the CO).
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Answer:

757,93 feets

Explanation:

We can make a right triangle between the boat (A), the Coast Guard officer (B) and the base of the observation tower (C), like in the graph attached. Now, you could also made a rectangle, adding the horizontal at the height of the Coast Guard, starting in B and ending in D, the vertex opossing C.

The angle of depression, its O in the graph.

Now, as we got an rectangle, of course, the segment AD its the same length as CB, and CA, the distance from the boat to shoreline, its the same length as DB.

ADB its an right triangle, with AB, the hypothenuse, and BD and DA, the catheti (or <em>legs</em>).

Now, we know the lenght BC, the height of the tower, 53 feets, so we also know the lenght of DA. DA its the opposite cathetus to the angle O. We wish to know the length AC, equal to the lenght DB, the adjacent cathetus of the angle O.

Know, the trigonometric function that connects the adjacent cathetus with the opossite cathetus its the tangent.

tangent( O ) = \frac{opposite}{adjacent}

We can take that the angle O = 4 °, and knowing that the opossite cathetus its 53 feets, we got:

tangent( 4) = \frac{53 feets}{DB}

DB=  \frac{53 feets}{tangent( 4)}

DB=  757,93 feets

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