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lozanna [386]
3 years ago
15

What does the following program segment do? Declare Count As Integer Declare Sum As Integer Set Sum = 0 For (Count = 1; Count &l

t; 50; Count++) Set Sum = Sum + Count End For
Engineering
1 answer:
Troyanec [42]3 years ago
6 0

1225

<u>Explanation:</u>

This segment helps initialize sum as 0. The for loop is used to increment with every execution and it is added to the sum. The loop runs 49 times and every time the count is added to the sum. In short it is the sum of first 49 natural numbers i.e 1+2+3+......+49.

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What is the relative % change in P if we double the absolute temperature of an ideal gas keeping mass and volume constant?
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Answer:  100% (double)

Explanation:

The question tells us two important things:

  1. Mass remains constant
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(We can think in a gas enclosed in a  closed bottle, which is heated, for instance)

In this case we know that, as always the gas can be considered as ideal, we can apply the general equation for ideal gases, as follows:

  1. State 1 (P1, V1, n1, T1)  ⇒ P1*V1 = n1*R*T1
  2. State 2 (P2, V2, n2, T2) ⇒ P2*V2 = n2*R*T2

But we know that V1=V2 and that n1=n2, som dividing both sides, we get:

P1/P2 = T1/T2, i.e, if T2=2 T1, in order to keep both sides equal, we need that P2= 2 P1.

This result is just reasonable, because as temperature measures the kinetic energy of the gas molecules, if temperature increases, the kinetic energy will also increase, and consequently, the frequency of collisions of the molecules (which is the pressure) will also increase in the same proportion.

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3 years ago
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Answer:

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Explanation:

Data:

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Calculations:

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ΔT = 75 °C - 25 °C = 50 °C  = 50 K

m = \text{100 lb} \times \dfrac{\text{1 kg}}{\text{2.205 lb}} \times \dfrac{\text{1000 g}}{\text{1 kg}}= 4.54 \times 10^{4}\text{ g}\\\\\begin{array}{rcl}q & = & mC_{\text{p}}\Delta T\\& = & 4.54 \times 10^{4}\text{ g} \times 4.18 \text{ J$\cdot$K$^{-1}$g$^{-1}$} \times 50 \text{ K}\\ & = & 9.5 \times 10^{6}\text{ J}\\ & = & \textbf{9500 kJ}\\\end{array}

2. Energy in British thermal units

\text{Energy} = \text{9500 kJ} \times \dfrac{\text{1 Btu}}{\text{1.055 kJ}} = \text{9000 Btu}

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