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yuradex [85]
3 years ago
13

A container with rigid walls is filled with 4.0 mol of air with Cv=2.5R Then the temperature is increased from 17 degrees C to 3

54 degrees C What is the change in internal energy? Let the ideal-gas constant R = 8.314 J/(mol • K).
337 J
28,000 J
7000 J
2800 J
Physics
1 answer:
galina1969 [7]3 years ago
3 0

Explanation:

Internal energy = heat + work

U = Q + W

Since there's no change in volume (rigid walls), W = 0.

U = Q

U = n Cᵥ ΔT

U = (4.0 mol) (2.5 × 8.314 J/mol/K) (354 C − 17 C)

U = 28,000 J

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Answer:

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3 years ago
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A crystalline grain of nickel in a metal plate is situated so that a tensile load is oriented along the [111] crystal direction
hoa [83]

Given:

Applied stress, \sigma = 0.45 MPa

Critical Resolved Stress,  T_{cRss}= 0.242 MPa

Solution:

a). According to the question, orientation of tensile load is along [1 1 1],

\sigma = 0.45 MPa

Now, for resolved shear stress,  \t_{Rss} along [1 0 1] within (1 1 T)

let  '\theta' be the angle between [1 1 1] and [1 0 1], then by coordinate formula:

cos\theta = \frac{1\times 1 + 1\times 0 + 1\times 1}{\sqrt{1^{2}+1^{2}+1^{2}\sqrt{1^{2}+0^{2}+1^{2}}}}

cos\theta = \sqrt{\frac{2}{3}}

let  '\phi' be the angle between [1 1 1] and [1  1 1], then by coordinate formula:

cos\phi  = \frac{1\times 1 + 1\times 1 + 1\times 1}{\sqrt{1^{2}+1^{2}+1^{2}\sqrt{1^{2}+1^{2}+1^{2}}}}

cos\phi = 1

Now, for the resolved components along [1 0 1]

\t_{Rss}  = \sigma  cos\theta cos\phi

\t_{Rss} = 0.45\times 10^{6}\times \sqrt{\frac{2}{3}}\times 1 = 0.3673 MPa

b).  For required tensile stress to produce  T_{cRss}= 0.242 MPa:

\sigma _{1} = \frac{T_{cRss}}{cos\theta  cos\phi }

\sigma _{1} = \frac{0.242\times 10^{6}}{\sqrt{\frac{2}{3}}\times 1}

\sigma _{1} = 0.2964 MPa

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3 years ago
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The electric output of a power plant is 716 MW. Cooling water is the main way heat from the powerplant is rejected, and it flows
Stels [109]

Answer:

(a) 83475 MW

(b) 85.8 %

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Output power = 716 MW = 716 x 10^6 W

Amount of water flows, V = 1.35 x 10^8 L = 1.35 x 10^8 x 10^-3 m^3

mass of water, m = Volume  x density = 1.35 x 10^8 x 10^-3 x 1000

                                                               = 1.35 x 10^8 kg

Time, t = 1 hr = 3600 second

T1 = 25.4° C, T2 = 30.7° C

Specific heat of water, c = 4200 J/kg°C

(a) Total energy, Q = m x c x ΔT

Q = 1.35 x 10^8 x 4200 x (30.7 - 25.4) = 3 x 10^12 J

Power = Energy / time

Power input = P = \frac{3 \times 10^{12}}{3600}=8.35 \times 10^{8}W

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(b) The efficiency of the plant is defined as the ratio of output power to the input power.

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7 0
3 years ago
4. How does light travel?
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4 years ago
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